Analiza Matematike 2 Integrali i Pacaktuar a.sh .Shabani

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1 Integrali i pacaktuar 1. Integralet tabelare Duke zbatuar tabelën e integrale të zgjidhen integralet: 1.

 (x

2.



3.



4.



5.

 1  x 

6.

4

7.

8x 2  2  2x  1 dx.

2

 3x  4)dx .

x 3 dx . x4 dx 3

x2

.

x  3 x dx . 

1

16  x x

x x dx .

dx .

Zgjidhja. 1.

2 2  ( x  3x  4)dx   x dx   3xdx   4dx 



2.



x 3 dx  x4

x

x

4

x3  3 x dx  4  dx 3

x3 x2 3  4 x  C. 3 2

dx  

3 dx dx dx   3  3 4   x 3 dx  3 x 4 dx 4 x x x

INTEGRALI I PACAKTUAR

2

x 31 x 4 1 1 1 3   2  3  C. 3  1 4  1 2x x



3.



dx 3



x2



dx x

2 3

2  1

2 3

1

x 3 x3   x dx    3 3 x  C. 2 1  1 3 3 

1 3

2

2 3



5.

1 1 1     1  x  x x   1  x  x 2 x 4 dx   1  x  x 4 dx

x  x dx 

x  x dx 

 1

3 4

  x dx   x 7

1

x3 3 x dx   x dx   3 x 5  C. 2 5 1 3

4.

3



4 3

1

1

3

3

3 4

1

3

1 x4  x dx    x 4 dx 3 1 4

3

4

7

x4 x4 4 4 4 3    x  x  C. 7 3 7 3 4 4

6. Mënyra e parë. 16  x

4

x

dx 



4 2  ( x )2 4 x

dx 



(4  x )(4  x ) 4 x

dx   (4  x ) dx

3

1 2

x2 2  4  dx   x dx  4 x   4x  x 3  C. 3 3 2

Mënyra e dytë. Kryjemë racionalizimin e emëruesit: 16  x

4

x

dx  

16  x



4 x

4 x 4 x

 4x 

2 x 3  C. 3

dx 



(16  x )(4  x ) dx   (4  x )dx 16  x

ANALIZA MATEMATIKE II

7.

3

8x 2  2 2(4 x 2  1) (2x  1)(2x  1) dx   2x  1  2x  1 dx  2 (2x  1) dx  2 (2x  1)dx





 2 2 x dx   dx  4

x2  2x  2x 2  2x  C. 2

Detyra plotësuese Të njehsohen integralet: 1.  ( x 3  x  2) dx. dx

4.

x

7.

 ( x  1) dx.

.

4

3



10.

  x

12.

x4 1  x 2 dx.

14.

2

x

dx .

Të njehsohen integralet: 8.

ex  x  e 1     x  dx.

9.

 e

2 x

10.

 (e

11.

e

12.

 (2

13.

2

x



 2x dx .

 x e ) dx .

e3 x  8 dx . x 2

x

x

x

.

6.

x

8.  ( x 2  1)2 dx .

9.

x



5.



 x x  x2  3 x 

9 x

3

3.

2.

 3x ) dx .  3 x dx.

x ( x  1) dx. dx 4

x

1   dx . x

13. 15.

11.





dx . 3x  1

3

x

x3 dx.

dx . x

x  2x  3

( x  1)( x 2  1) dx .  x3 9x 2  3

4

3

(1  x 2 ) dx .

dx .


4

14.

 (3

15.

3x 1  5x 1  15x dx.

x

 4 x )2 dx .

Zgjidhja. 8.

ex x  e 1    x

9.



 dx x  e x  ln| x | C.  dx   e dx   x 



e 2 x  2x dx   ( e 2 x  2x ) dx   e 2 x dx   2x dx   e 2 e x dx   e 2  e x dx 

2x ln 2

2x 2x 2x  e2 e x   e 2 x   C. ln 2 ln 2 ln 2 x e 1  C. e 1

10.

x e x e x  (e  x ) dx   e dx   x dx  e 

11.

e3 x  8 ( e x )3  23 ( e x  2)( e 2 x  2e x  4) dx  dx   ex  2  ex  2  ex  2   e 2 x dx  2 e x dx  4  dx   ( e 2 )x dx  2e x  4 x 

( e 2 )x e2x e2x x x  2 e  4 x   2 e  4 x   2e x  4 x  C. 2 ln e 2 ln e 2

12.  (2x  3x ) dx   2x dx   3x dx 

2x 3x   C. ln 2 ln 3 x

2 x 3 2 13.  2x  3 x dx   2x (31 )x dx     dx     C. 2 3 ln 3

14.

 (3

x

 4 x )2 dx   (32 x  2  3x 4 x  4 2 x )dx   9 x dx  2 12x dx   16 x dx 

9x 12x 16 x 2   C. ln 9 ln12 ln16 x

15.

x

3x 1  5x 1 3  3x  51  5x 1 1  1  dx  dx  3   dx     dx  15x  5 5 3 15x  


5 x

x

1  1  5 1 3 3 x 1 3       5  3  x  C. 1 5 1 ln 5 5 ln 5 ln ln 5 3

Detyra plotësuese Të njehsohen integralet:  e x  16.  e 1  2  dx. x   x

 x  1  1  x 2  17.   e 2     dx .   2   

18.

2x 2e  (e  x ) dx.

19.

e4 x  1  e2x  1dx.

20.

e3 x  27  e x  3 dx.

21.

3

22.

 (2

23.

 (3

24.

2x 1  3x 1  6x dx.

25.

 (e

x

 2 x ) dx .


x2  x 2  1 dx.

17.

x4  1  x 2 dx.

18.

(1  x )2  x (1  x 2 ) dx.

19.

x2  4  x 2  1 dx.

20.

x3  x  2  x 2  1 dx.

21.

  4e

 

x



  dx . 4  4x 2  5

 5 x dx .

x

x

 3 x )2 dx .

x e

 x e 1 ) dx .


6

Zgjidhja. 16. Mënyra e parë. Pjesëtojmë polinomit x 2 me polinomin x 2  1. Merret: x2 1 1 2 . x2  1 x 1

Pra, x2 1  dx   x 2  1 dx   1  x 2  1  dx   dx   x 2  1  x  arctan x  C.

Mënyra e dytë. x2  x 2  1 dx 

 x2  1 x2  1 1 1  dx   x2  1   x 2  1  x 2  1  dx

1     1  2 dx  x  arctan x  C. x  1  

17.

18.

x4  1  x 2 dx 

x4 1  1 x4 1 1 dx   1  x2  1  x 2 dx   1  x 2 dx



( x 2  1)( x 2  1) dx  arctan x   ( x 2  1)dx  arctan x  1  x2



x3  x  arctan x  C. 3

(1  x )2 1  2x  x 2 dx   x (1  x 2 )  x (1  x 2 ) dx 



19.

x2  4  x 2  1 dx 



1  x2 2x  x (1  x 2 ) dx   x (1  x 2 ) dx

dx dx  2  ln| x | 2 arctan x  C. x 1  x2

x2  1  3 3  dx   x 2  1 dx   1  x 2  1  dx   dx  3 x 2  1

 x  3 arctan x  C.

20. Pas pjesëtimit të polinomeve x 3  x  2 me x 2  1 merret: x3  x  2 2 x 2 . 2 x 1 x 1


7

Kemi: x3  x  2 2  x2  dx  x  dx   2 arctan x  C.  x2  1   x 2  1  2

21.



  4e

x





 dx 5 dx  4  e x  5  4e x  arcsin x  C.   2 4  4x 2  2 1  x2 5

Detyra plotësuese Të njehsohen integralet: 26.

3x 2  1  x 2  1 dx.

27.

x4 1  x 2  1 dx.

28.

(1  x )3  x 2  x 3 dx .  x (1  x )

29.

x3  x  3  x 2  1 dx.

x8 1  x 2  1 dx.

   31.   3x    

30.

   1  dx . 2 1 x   9  3  


 ( x  sin x  2 cos x ) dx.

23.

 sin

2

24.

 sin

2

25.

 tan

26.

 cos

27.



cos 2x dx . x  cos2 x

2

2

dx . x  cos2 x x dx .

x dx . 2

1  sin 2x dx .

Zgjidhja. 22.

 ( x  sin x  2 cos x ) dx   x dx   sin x dx  2 cos x dx


8



x2  (  cos x )  2 sin x  C 2 

x2  cos x  2 sin x  C. 2

23. Zbatojmë formulën cos 2x  cos2 x  sin 2 x.

Merret: cos 2x cos2 x  sin 2 x dx   sin2 x  cos2 x  sin2 x  cos2 x dx

24.



cos2 x sin 2 x dx   sin2 x  cos2 x  sin2 x  cos2 x dx



 sin

dx 2

x



dx   cot x  tan x  C. cos2 x

dx sin 2 x  cos2 x dx dx   sin2 x  cos2 x  sin2 x  cos2 x dx   cos2 x   sin 2 x  tan x  cot x  C.

25.

2  tan x dx 

sin 2 x 1  cos2 x dx dx   cos2 x  cos2 x dx   cos2 x   dx

 tan x  x  C.

26. Meqë cos2 x 

1  cos 2x x 1  cos x atëherë cos2  . 2 2 2

D.m.th.

 cos 27.



2

x 1  cos x 1 1 1 1 dx   dx   dx   cos x dx  x  sin x  C. 2 2 2 2 2 2

1  sin 2x dx 



sin 2 x  cos2 x  2 sin x cos x dx

 (sin x  cos x )  |sin x  cos x |dx   (sin x  cos x )  sgn(sin x  cos x ) dx 

2

 (  cos x  sin x ) sgn(sin x  cos x )


9

 (sin x  cos x ) sgn(sin x  cos x )  (sin x  cos x ) sgn(cos x  sin x ).

Detyra plotësuese Të njehsohen integralet: x  32.    2sin x  3cos x  dx. 2 

4

34.

 sin

36.

2  sin

38.

1  sin3 x  sin3 x dx.

2

2x

dx .

x dx . 2

cos 2x dx . 2 2x

33.

 sin

35.

 cot

37.

x x    sin 2  cos 2  dx.

2

x dx . 2

2. Integrimi me metodën e zëvendësimit Duke zbatuar metodën e zëvendësimit të njehsohen integralet: 1.

 (1  x ) dx. 7

2.

 (3  2x ) dx. 6

3.



1  xdx .

4.



4

1  x dx .

Zgjidhja. 1.

7  (1  x ) dx 

1x t t8 (1  x )8   t7 dt   C   C. 8 8 dx  dt

3  2x  t 1  1 2.  (3  2x ) dx  2dx  dt   t 6   dt    t 6 dt 2  2 1 dx   dt 2 6



1 t7 t7 (3  2x )7 C   C   C 2 7 14 14

3. Mënyra e parë: 1x t



1  x dx  dx  dt  dx  dt



1

1 2

1

t2 t ( dt )    t dt   C 1 1 2

2

2 2   t3  C   (1  x )3  C. 3 3

Mënyra e dytë: 1  x  t2



1  x dx 

2   t( 2t )dt  2 t 2dt   t 3  C 3 dx  2tdt dx  2tdt

t  1x



4.



4

1  xdx 

2 (1  x )3  C. 3

1  x  t4  t  4 1  x 3

dx  4t dt

3

  t  4t dt  4  t 4 dt


2

 4

t5 4  C   4 (1  x )5  C. 5 5

Detyra për ushtrime Të njehsohen integralet: 4

1.

 (1  2x ) dx.

2  2.    x  dx . 3 

4.



5.

7.

 ((1  x )

5

3

1  3x dx . 3



5

5x  1 dx .

3.



6.

(

1

x dx . 2

x  1  x  1)dx .

 3 1  x )dx .


2x  1  x 2  x  3 dx.

8.

x

2

 1  x 3 dx .

x3  x 4  1 dx. x2  3 dx . 9.  3 x 2

6.

7.

x

1  2x 2 dx .

Zgjidhja. 5.

6.

x2  x  3  t 2x  1 dt dx    ln|t | ln( x 2  x  3)  C.  x2  x  3 t (2x  1)dx  dt

x

x

3

4

1

x4  1  t dx  4 x 3 dx  dt  1 x 3 dx  dt 4



1 dt ln( x 4  1) 1 dt 1 4  C.    ln|t |  4 t 4 t 4

1  2x 2  t 2  t  1  2x 2 1 7.  x  1  2x 2 dx  4 xdx  2tdt   t  t dt 2 1 xdx  tdt 2 3 1 t 1    (1  2x 2 )3  C. 2 3 6


3

1  x 3  t2  t  1  x 3

8.



1  x 3  x 2 dx  3x 2 dx  2tdt x 2 dx 



9.



x2  3 3

x 2

dx 

2 tdt 3

2 2 t3   t  tdt    C 3 3 3

2 (1  x 3 )3  C 9

x  2  t 3  x  t 3  2; t  3 x  2 dx  3t 2 dt

(t 3  2)2  3  3t 2 dt  3 (t 6  4t 3  4  3)  t dt  t t8 t5 t2  3 (t7  4t 4  t )dt  3   12   3  8 5 2 33 12 3 3  ( x  2)8  ( x  2)5  3 ( x  2)2  C. 8 5 2 

Detyra plotësuese Të njehsohen integralet 8.

2x  1  x 2  x  4 dx.

9.

2x  3  x 2  3x  1 dx.

10.

x5  x 6  1 dx.



11.

2x  a 2x 3 , a-const.12. dx  x 2  ax  1  4x 4  1 dx.

13.

14.

2  x  1  x dx.

15.

2  x  2  3x dx.

16.

17.

2 3  x  2  5x dx.

18.

 x 3   x  5  

  dx . x 2 x2 2

Të njehsohen integralet: dx

10.



x 1  x 1

12.



1x dx . 1x

.

1x

11.



13.

x



1  x2

dx .

dx x2 1

.

x 2 x 3

dx .

x ( x 2  1) 3

x2  1

dx .


4

Zgjidhja. 10. Pas racionalizimit të emëruesit kemi: dx 1 x 1  x 1 I    dx x 1  x 1 x 1  x 1 x 1  x 1 x 1  x 1 x 1  x 1  dx   dx x  1  ( x  1) 2 1 1  x  1dx   x  1dx  ( I1  I 2 ) ,  2 2 ku I1   x  1 dx ; I 2   x  1 dx .





Zgjidhim veçmas integralet I1 , I 2 . Merret: x  1  t2

I1 



t3 2 x  1dx  t  x  1   t  2tdt  2   ( x  1)3  C. 3 3 dx  2tdt x  1  t2

I2 



t3 2 x  1dx  t  x  1   t  2tdt  2   ( x  1)3  C. 3 3 dx  2tdt

Pra





12 2 1  ( x  1)3  ( x  1)3   C  ( x  1)3  ( x  1)3  C.  23 3 3  1x dx xdx 11. I   dx     arcsin x  I1 . 2 2 1x 1x 1  x2 I 

I1 



1  x 2  t2  2xdx  2tdt  1  x2 xdx  tdt xdx



tdt    dt  t   1  x 2 t

Pra, I  arcsin x  1  x 2  C 12. Pasi të kryejmë racionalizimin merret 1x 1x 1x 1x  1  x dx   1  x  1  x dx   1  x 2 dx. Integrali që morëm është i ngjashëm me integralin e detyrës paraprake. 1x Pas zgjidhjes merret  dx  arcsin x  1  x 2  C. 1x


13.

x 



dx x2 1





5

dx 1   x x 2 1  2  x  

dx 1 x  x sgn x 1    x

2







dx 1 x |x | 1    x dx

1  sgn x 

1 x2 1    x

2

2

1 t dx dt x  sgn x     sgn x  2 1 1  t2 1 2   dt x 1  x2 x 1   sgn x  arcsin t  C   sgn x  arcsin  C. x

Detyra për ushtrime Të njehsohen integralet: 19.



22.



25.



dx

dx

20.



x 3 dx . 1  (2x )2

23.



1x dx . 1x

26.



15.

x

18.

1 x

x 1  x 1

.

1

21.



dx .

24.



.

27.

x

16.

 x ln x ln(ln x ) .

2x  1  2x  1

x 1 1  9x

2

xdx 4

x x

2

x 2 dx . 1  x2

3

.

1x 1  4x 2 dx x2  1

dx . .

Të njehsohen integralet: dx

14.

 x (1  2 ln x ).

17.

 x cos

19.

dx



2

ln x

.

ln( x  1  x 2 ) dx . 1  x2

dx 2

1  ln x

1

2

ln

.

1x dx. 1x

dx


6

Zgjidhja. (1  2 ln x )  t dx 1 dt 1 14.   dx 1    ln|t | x (1  2 ln x ) 2 t 2  dt x 2 1  ln|1  2ln x |C. 2 ln x  t dx dt 15.   dx   arcsin t  C  arcsin(ln x )  C. 2  dt x 1 - ln x 1  t2 x ln(ln x )  u dx du 16.   dx   ln|u | C  ln|ln(ln x )| C. x ln x ln(ln x ) u  du x ln x ln x  t dx dt 17.   dx   tan t  C  tan(ln x )  C. 2 x cos ln x cos2 t  dt x 1 1x 18. I   ln dx 2 1x 1x 1x Zëvendësojmë ln u 1x Pas diferencimit merret: '

1 1  x   dx  du 1  x  1  x  1x 1  x (1  x )' (1  x )  (1  x )(1  x )'  dx  du 1x (1  x )2 1 1  x  (1  x )( 1)  dx  du 1x 1x 1  x 1  x dx  du 1  x2 1 1 dx  du 2 1x 2 Pra, kemi: 2

 1x   ln 1  x  2 du u   1 ln 2 1  x  C. I  u   2 4 4 4 1x




19. I 

7

ln( x  1  x 2 ) ln( x  1  x 2 ) dx  dx  1  x2 1  x2

Zëvendësojmë ln( x  1  x 2 )  u Pas diferencimit merret: ( x  1  x 2 )'

dx  du x  1  x2 x 1 1  x 2 dx  du x  1  x2

x  1  x2 1  x2

dx  du x  1  x2 dx Pra  du. 1  x2 Merret: I 



3

1 2

u2 2 udu   u du   ln3 ( x  1  x 2 )  C. 3 3 2

Detyra për ushtrime Të njehsohen integralet: dx

28.

 2x (ln x  1).

31.

34.

dx

29.

x

 ln(sin x ) dx.

32.

 cos

1 1x  1  x 2 ln 1  x dx.

35.

cos x



1  ln x 2

.

dx . x ln(ctgx )

sin x

30.

 ln(cos x ) dx.

33.

 x sin

dx 2

ln x

.

ln( x  1  x 2 ) dx . 1  x2


cos x  e sin x dx.

21.

ex  1  e x  x dx. 22.



ex 1 e  2 x

dx . 23.

x

2  x3

e

dx .


8 1

ex 24.  2 dx . x

25.

dx  ex  ex .

26.



ex ex  1

dx .

Zgjidhja. cos x  t

20.

e

21.

ex  x  t ex  1 dt dx    ln|t | ln| e x  x | C.  ex  x x t ( e  1)dx  dt

22.



cos x

sin xdx 

ex ex 

1 2

dx 

 sin xdx  dt

1 t dt 1    ln|t | ln  e x    C. 2 t 2  e x dx  dt ex 

x 3  t

23.

x

2  x3

e

   et dt  et  e cos x  C.

dx  3x 2 dx  dt   1 x 2 dx   dt 3

1 t 1 1 3 e dt   et   e  x  C.  3 3 3

1 1 t 1 ex x    et dt  et  e x  C. 24.  2 dx  1 x  2 dx  dt x dx dx dx e x dx 25.  x     e2x  1  e2x  1 e  ex  x 1 e  x e ex 

26.



ex x

e 1

ex  t e x dx dt   2  arctan t  arctan e x  C. x 2 x ( e )  1 e dx  dt t 1

dx 

ex  1  t2  t  ex  1 x

e dx  2tdt





2tdt  2 e x  1  C. t


sin x  e cos xdx.

37.

e tgx  cos2 x dx.

38.

e

dx . 1

x


39. 42.

9

2e 2 x  1  e2x  x dx.



e x dx 3

ex  3

40.

x

2 2 x 3

e

dx .

41.

 xe

x 2 a2

dx .

.

Të njehsohen integralet: dx 8x  3 x 2 dx 27.  2 28. 29.  2 . dx . . 2 6  x a x 7 9x Zgjidhja. dx x 2 t dx 1 dx 1 a  2  2  2 27.  2 a 2 2 2 x a x a a x a  1 dx  adt 2   a a 1 dt 1 1 x   2  arctan t  arctan  C. a t 1 a a a

t

adt 2 1

dt x dx x dx 1 dt  2  3x 2 dx  dt   2 3 2   2 2 . 28. I   6 3 2 3 3 t 9x 3  (x ) 3 t dt 2 x dx  3 Në bazë të detyrës paraprake dt 1 t 1 x3  arctan  C  arctan  C1 . 1  32  t 2 3 3 3 3 1 x3 Pra I  arctan  C. 9 3 8x  3 8x dx 29.  2 dx  8  2 dx  3 2  8 I1  3 I 2 , ku x 7 x 7 x  ( 7 )2 xdx dx I1   2 ; I2   2 x 7 x  ( 7 )2 2

2

x3  t

x2  7  t I1 

Pra

xdx 1 dt 1 1  2xdx  dt    ln|t | ln( x 2  7). 2 2 t 2 2 7 dt xdx  2

x


10

1 1 x ln( x 2  7)  3 arctan C 2 7 7 3 x  4 ln( x 2  7)  arctan  C. 7 7

I  8

Detyra për ushtrime Të njehsohen integralet 43.

dx  2x 2  a 2 .

46.

 2x

x2 dx . 6 5

44.

dx  2x 2  3 .

47.

x

9x  1 dx . 2  15

45.

x3  7  x 8 dx.

48.

 (ex )

dx . 2  2


a arctan x  1  x 2 dx , a  R \ {1}, a  0.

31.

dx

 arcsin x

1  x2 e arctg2 x  x 33.  dx . 1  4x 2

.

arctanx  x ln(1  x 2 ) dx .  1+x 2 Zgjidhja. arctan x  u a arctan x au a arctan x u 30.  dx   a du    C. dx  ln| a | ln a 1  x2  du 2 1x arcsin x  u dx du 31.     ln|u | ln|arcsin x | C. dx 2  du u arcsin x 1  x 1  x2

32.

arctan x  x ln(1  x 2 ) arctan x x ln(1  x 2 ) dx  dx    1  x2  1  x 2 dx 1  x2  I1  I 2 ,

32. I 

arctan x  u u2 (arctan x )2  u du   . dx  2 2  du 2 1x ln(1  x 2 )  u ln(1  x 2 ) 1 u2 ln 2 (1  x 2 ) I2   dx   udu   . 1 du  2 4 4 1  x2 xdx  2 1  x2 Pra

ku I1 

arctan x  1  x 2 dx 


11

(arctan x )2 ln 2 (1  x 2 )   C. 2 4 e arctan 2 x  x e arctan 2 x xdx 33. I   dx  dx    I1  I 2 , 2 2  1  4x 1  4x 1  4x 2 I 

arctan 2x  u I1 

I2 

arctan 2 x

e 1 1 u eu e arctan 2 x dx   2 dx  du  e du   . 2 2  1  4x 2 2 2 1  4x dx 1  du 2 2 1  4x 1  4 x 2  t 1 dt 1 xdx 1 2   1  4x 2 8xdx  dt  8  t  8 ln|t | 8 ln(1  4x ).

Pra, I 

e arctan 2 x 1  ln(1  4 x 2 )  C. 2 8


3arctan x  1  x 2 dx.

50.

52.

x ln(1  3x 2 )  1  3x 2 dx.

53.

Të njehsohen integralet: 34.  sin ax dx. 35. 37. 40. 43.

 cot ax dx.  tan x dx.  sin x sin 2x dx. 4

51.



3

arcsin 2 x 1  x2

dx .

e arctan 4 x  e arctan 4 x dx .  1 x2  16

 cos ax dx.

36.  tan ax dx .

cos 2x

 sin x cos x dx. 39.  sin x dx. 41.  sin x dx . 42.  cot x dx . 44.  cos x cos 2x cos 3x dx. 38.

dx  2 sin2 x  cos2 x . 46. sin x  sin 3 x 48.  dx . 49. 2 cos2 x  sin 2 x

45.

arctan 3 x  1  9x 2 dx.

4

5

sin 2x  tan4 x dx. dx  sin x .

5

47.

sin 3 x  cos2 x dx.

50.

 sin

4

x  cos4 x dx .


12

Zgjidhja. ax  t

34.

 sin ax dx  adx  dt dx 



1 1 1 sin t dt   cos t   cos ax  C.  a a a



1 1 1 cos t dt  sin t  sin ax  C.  a a a

1 dt a

ax  t

35.

 cos ax dx  adx  dt dx 

1 dt a

cos ax  u sin ax 36.  tan ax dx   dx  1 cos ax sin ax dx   du a 1 du 1 1     ln|u|  ln|cos ax |C. a u a a sin ax  u cos ax 37.  cot ax dx   dx  1 sin ax cos ax dx  du a 1 du 1 1    ln|u| ln|sin ax |C. a u a a 38. Mënyra e parë: cos 2x cos2 x  sin 2 x cos x sin x I  dx   dx   dx   dx sin x cos x sin x cos x sin x cos x   cot x dx   tan x dx  ln|sin x |  ln|cos x |  ln|sin x  cos x |C.

Mënyra e dytë: cos 2x 2 cos 2x cos 2x I  dx   dx  2 dx  2 cot 2x dx sin x cos x 2 sin x cos x sin 2x  ln|sin 2x | C1 . Shënim.


13

Lexuesi mund të ketë përshtypjen se rezultatet që morrëm nga zgjidhja në dy mënyrat janë të ndryshme por lehtë vërejmë se: ln|sin 2x | C1  ln|2 sin x cos x | C1  ln 2  ln|sin x cos x | C1  ln|sin x cos x | C. 2

 1  cos 2x  4 2 2  sin xdx   (sin x ) dx    2  dx 1   (1  2 cos 2x  cos2 2x ) dx 4 1 1 1 1  cos4 x   dx   cos2x dx   dx 4 2 4 2 1 1 1  x  I1  I 2 . 4 2 4 2x  u 1 1 1 I1   cos 2x dx    cos u du  sin u  sin 2x  C1 . 1 2 2 dx  du 2 2 1  cos 4 x 1 1 1 1 I2   dx   dx   cos 4 x dx  x  sin 4 x  C2 . 2 2 2 2 8 Përfundimisht merret: 1 1 1 1  1 1  3x sin 2x sin 4 x I  x   sin 2x    x  sin 4 x      C. 4 22 4 2 8 8 4 32     1   1  dx 40.  tan 4 x dx   tan 2 tan 2 x dx   tan 2 x  2  cos x  1   tan 2 x dx   tan 2 x dx  I1  I 2 . cos2 x tan x  u 1 u3 tan3 x 2 2 I1   tan x dx   u du   . dx 3 3 cos2 x  du  2 cos x 2 sin x 1  cos2 x dx I 2   tan 2 x dx   dx   dx    dx 2 2 cos x cos x cos2 x   tan x  x  C. Përfundojmë se tan3 x 4 tan x dx   tan x  x  C.  3 cos x  t 41.  sin 5 xdx   sin x sin 4 x dx   sin x (1  cos2 x )2 dx  sin xdx  dt

39.


14

   (1  t 2 )2 dt    dt  2 t 2 dt   t 4 dt  t  2

t3 t5  3 5

2 1 cos3 x  cos5 x  C. 3 5 5 cos x cos x  cos4 x cos x (1  sin 2 x )2 42.  cot 5 xdx   dx  dx  dx  sin5 x  sin5 x sin 5 x sin x  u (1  u2 )2 1  2u2  u4   du  du  cos x dx  du u5 u5 du 1 1   u 5 du  2 u 3 du     4  2  ln|u | u 4u u 1 1    ln|sin x | C. 4 sin 4 x sin 2 x Në detyrat vijuese zbatohen formulat: 1 sin x  sin y  (cos( x  y)  cos( x  y)) 2 1 cos x  cos y  (cos( x  y )  cos( x  y )) 2 1 sin x  cos y  (sin( x  y)  sin( x  y)). 2 1 43.  sin x sin 2x dx   (cos( x  2x )  cos( x  2x )) dx 2 1 1 1  cos x dx   cos 3x dx  sin x  sin 3x  C.  2 2 6 1 44.  cos x cos 2x cos 3x dx   cos 2x  (cos( x  3x )  cos( x  3x )) dx 2 1   cos 2x (cos( 2x )  cos 4 x ))dx 2 1  cos 2x cos 2x dx   cos 2x cos 4 x dx 2  1  dx   cos 4 x dx   cos 2x dx   cos 6x dx 4  1 1 1 1    x  sin 2x  sin 4 x  sin 6 x   C. 4 2 4 6  dx dx tan x  u 2 2 dx cos x cos x 45.     dx 2 sin 2 x  cos2 x  sin 2 x 2 tan 2 x  1  du 2 1 cos2 x 2 cos x   cos x 



 








15



du du   2 arctan 2u 2 1 ( 2u )2  1

 2u

 2 arctan( 2 tan x )  C. sin 2x 2 sin x cos x sin x cos x cos4 x 46.  dx   dx  2 dx 4 4 tan x sin x sin 4 x cos4 x sin x  u sin x (1  sin 2 x )2 cos x  2 dx  4 cos x dx  du sin x u(1  u2 )2 du du du du  2 3  4  2  2 4 u u u u 1 4 1 4   2   2 ln|u |    2 ln|sin x | C. u u sin 2 x sin x sin3 x sin x sin 2 x sin x (1  cos2 x ) 47.  dx  dx  dx  cos2 x  cos2 x cos2 x  2



cos x  u sin x dx  du

 

1  u2 1 1 du   u   cos x  C. u cos x u2

sin x  sin 3 x sin x (1  sin 2 x ) sin x cos2 x dx   dx   dx 2 2 2 x  sin x cos x  1 cos2 x  1 cos x  t t2 dt    2 dt    dt   2 sin x dx  dt t 1 t 1  t  arctan t   cos x  arctan(cos x )  C. x x sin 2  cos2 dx dx 2 2 dx   49.  x x  x x sin x  2 sin cos 2 sin cos 2 2 2 2 x x   sin cos  1 1 2 2   dx   dx   ( I 2  I 2 ). x x 2  2 sin  cos 2 2  

48.

 2 cos


16

x u 2 x sin 2 dx   1 sin x dx  du  du  2 ln|u |  2 ln|cos x |. I1   u x 2 2 2 cos x 2 sin dx  2 du 2 cos

x x sin  u du x 2 2 dx  I2    2  2 ln|u | 2 ln sin . x u 2 x sin cos dx  2du 2 2 x sin 1 x x  2  ln tan x  C. Pra I   2 ln cos  2 ln sin   ln x 2 2 2 2 cos 2 cos

4

 2 sin x cos x  50.  sin x  cos x dx   (sin x cos x ) dx     dx 2   4

4

4

2

(sin 2 2x )2 1  1  cos 4 x   16 dx  16   2  dx 1  (1  2 cos 4 x  cos2 4 x ) dx 64  1  dx  2 cos4 x dx   cos2 4 x dx 64  1  1 1  cos 8x   x  2  sin 4 x   dx   64  4 2  







1  1 1 1  x  sin 4 x  x  sin 8 x   64  2 2 16 



1 3 1 1  x  sin 4 x  sin 8 x   C.  64  2 2 16 

Detyra për ushtrime Të njehsohen integralet: cos x dx . x x sin cos 2 2

54.



57.

 cos

5

x dx .

55.

 cos

58.

 tan xdx.

4

5

x dx .

56.

 cot

59.

 sin 3x sin x dx.

4

x dx .


17

dx . x  5 cos2 x

60.

 sin x sin 2x cos 3xdx.

61.

 sin

62.

cos 2x  cot4 x dx.

63.

sin 2x  cot4 x dx.

65.

cos2 x  sin 2 x  sin x  sin3 x dx.

66.

 cos x .

Të njehsohen integralet dx 51.  . 2 x  a2 53.



a 2  x 2 dx .

2

64. 

dx

52.

x

54.



67.

dx a2  x 2 dx (1  x 2 )3

sin5 x dx . cos2 x



dx . x sin 2

. .

Zgjidhja. x 2  a2  t  x , t  x  x 2  a2 x 2  a 2  t 2  2tx  x 2

51.

dx





x 2  a2



t2  a2 t2  a2 , dx  dt 2t 2t 2 t2  a2 x 2  a2  t  2t 2 t  a2 x 2  a2  2t

 2tx  t 2  a 2  x 

t2  a2 2t 2 dt  dt  ln|t | ln| x  x 2  a 2 | C. t t2  a2 2t

x  a tan u dx du 52.   dx  a 2 2 cos2 u x a x a2  x 2  a

du 2 cos u  2 a tan u a  a 2 tan 2 u a

1 cos u


18

du du 2 1 du cos u cos2 u     2 1 a sin u a tan u 1  tan u a sin u cos u 1 u  ln tan . a 2 a x Meqë cos u  atëherë sin u  dhe 2 2 2 a x a  x2 x tan

Pra

53.



u sin u   2 1  cos u

x

dx 2

a x

2



1

a2  x 2 a



x a  a2  x 2

.

a2  x 2

1 x ln  C. a a  a2  x 2

x  a sin t dx  a cos t dt a 2  x 2 dx  sin t  x  a x t  arcsin a



a 2  a 2 sin 2 t  a cos t dt

 a 2  1  sin 2 t cos t dt  a 2  cos2 t dt  a 2  



 dt   cos 2tdt  a2  t  12 sin 2t   2

1 x x x2    a2    t  2 sin t cos t  arcsin  1    2  2 a a a 2     a2 x x  arcsin  a 2  x 2  C. 2 a 2 dt x  tan t dx dt cos2 t    dt 3 dx  (1  x 2 )3 (1  tan 2 t )3  sin 2 t  2 cos2 t cos t 1   cos2 t   

54.

a2 2

1  cos 2t dt 2

a2 2








19

dt cos2 t

1 cos3 t

x 1  x2

  cos t dt  sin t  C 

tan t 1  tan 2 t

 C.

Shënim. sin 2 t sin t tan 2 t tan t cos2 t sin 2 t     sin t  . 2 2 2 2 2 2 sin t  cos t sin t cos t 1  tan t 1  tan t  cos2 t cos2 t 2




71.



dx 2

x a

69.

2

dx ( a 2  x 2 )3

.

x

dx 2

x 1

.

70.



dx 9  x2

.

3. Integrimi me metodën parciale Të njehsohen integralet: 1.

 ln x dx.

2.

 x ln x dx.

3.

x

n

ln x dx .

4.

x

4

ln3 x dx .

Zgjidhja. ln x  u dx  du 1.  ln x dx  x

 x ln x   x 

dx  x ln x   dx x

 dx  v  v  x  x ln x  x  C.

ln x  u

2.

 x ln x dx 

dx  du x v   xdx ; v 





x2 x 2 dx ln x    2 2 x

x2 2

x2 1 x2 x2 ln x   xdx  ln x   C. 2 2 2 4

ln x  u dx  du 3.  x n ln x dx  x v   x n dx ; v 



x n1 x n1 dx ln x    n 1 n 1 x

x n1 n 1



x n1 1 x n1 1 x n1 n ln x  x dx  ln x  n 1 n 1  n 1 n  1 (n  1)



x n1 x n1 ln x   C. n 1 (n  1)2


2

ln3 x  u

4.

4 3  x ln x dx 



3 ln 2 x x5 3 3x 5 ln 2 x dx  du  ln x    dx x 5 5 x x5 v   x 4 dx  5

x5 3 3 x5 3 3 ln x   x 4 ln 2 xdx  ln x   I1 , 5 5 5 5

ku ln 2 x  u I1   x 4 ln 2 dx 



2 ln x x5 2 x 5 2 ln x dx  du  ln x   dx x 5 5 x x5 v 5

x5 2 2 x5 2 2 ln x   x 4 ln x  ln x  I 2 . 5 5 5 5

Në bazë të detyrës 3. I 2   x 4 ln xdx 

Pra I1 

x5 x5 ln x  2 5 5

x5 2 2  x5 x5  x5 2 2 5 2 5 ln x   ln x  2   ln x  x ln x  x . 5 5 5 25 125 5  5

Përfundimisht, I 

x5 3 3  x5 2 5 2 5 ln x   ln 2 x  x ln x  x   C. 5 5 5 25 125 


x

2

ln x dx .

2.

 x ln

2

x dx .

3.

x

3

ln 2 x dx .


3


ln 2 x  x 2 dx.

6.

 (x

2

7.

 ln( x 

8.

x

ln

1  x 2 )dx .

2

 x ) ln( x  1) dx .

1x dx. 1x

Zgjidhja. ln 2 x  u ln 2 x 1 1 1 1 5.  2 dx  2 ln x  dx  du   ln 2 x    2 ln x  dx x x x x x 2 1 v   x dx   x



1 2 ln x 1 ln x  2 2 dx   ln 2 x  2 I1 , x x x

ku ln x  u ln x dx 1 dx 1 I1   2 dx   du   x 1 ln x     ln x   x 2 dx x x x x x v   x 1



1 1 ln x  . x x

Pra, I 

1 2 1 1  1 ln x  2   ln x     (ln 2 x  2 ln x  2)  C. x x x  x

ln( x  1)  u  x3 x2  dx  du    6.  ( x 2  x ) ln( x  1)dx   ln( x  1) x 1 2   3 x3 x2 v  3 2

 x 3 x 2  dx  x3 x2  2x 3  3x 2      dx   ln( x  1)   2  x 1  3 2  6( x  1)  3


4

 x3 x2     ln( x  1)  I1 , 2   3

ku I1 



2x 3  3x 2 2x 3  2x 2  x 2 2 x 2 ( x  1) dx   dx   dx 6( x  1) 6( x  1) 6 x 1



1 x2 1 1  1  dx   x 2 dx    x  1  dx  6 x 1 3 6  x  1 



 x3 1  x2    x  ln( x  1)  . 9 6 2 

Pra  x3 x2  x3 x2 x 1 I      ln( x  1)  ln( x  1)   2  9 12 6 6  3  x3 x2 1  x3 x2 x     ln( x  1)     C. 2 6 9 12 6  3 ln( x  1  x 2 )  u ( x  1  x 2 )'

dx  du x  1  x2 1 1 2 x  1 dx  du 7.  ln( x  1  x 2 )dx  x  1  x2 dx  du x2  1 v   dx  v  x

 x ln( x  1  x 2 )  

xdx 2

x 1

 x ln( x  1  x 2 )  1  x 2  C.


5

1x u 1x 1x 2 x3 1  x x3 2 dx  2 dx  du  ln   2 dx 8.  x 2 ln 1x 3 1x 3 x 1 x 1 x3 v 3 ln



x3 1  x 2  x  x3 1  x 2 x2 ln  x  2 dx  ln    3 1x 3  3 1x 3 2 x 1 



2 1 x3 1  x x2 1  ln| x 2  1| ln   ln| x 2  1| C. 3 2 3 1x 3 3

Detyra për ushtrime Të njehsohen integralet:



4.

2

x ln 2 x dx .

5.

 ln x    x  dx.

6.



ln x dx . x3


 xe

x

dx .

10.

x e

3 x2

dx .

11.

e

x

dx .

12.

e

x  ln x

dx .

Zgjidhja. x u

9.

x  xe dx 

dx  du v   e  x dx

  xe  x   e  x dx   xe  x  e  x  C.

v  e  x x2  u 2xdx  du

10.

x e

3 x2

dx   x 2  x  e  x dx  v  xe  x 2 dx    2

v



2 1 2 x2 1 x e   2xe  x dx 2 2

1 x2 e 2

2 2 1 2 x2 1 1 2 1 2 x e   xe  x dx   x 2 e  x  e  x   e  x ( x 2  1)  C. 2 2 2 2


6

1

x u

11.

x  e dx 



 2 xe

x

x

x

e

x

  2e

dx 

x t e dx v dx   dt  2 et dt  2e x 2 x dx  2dt x

dx

x

dx  du

2 x

2 x

x



x

 2 xe

e

x

x

dx  2 xe

x

 2e

x

x

 2( x  1)e x .

12.

e

x  ln x

dx   e x e ln x dx   xe x dx 

x  u, dx  du ve

x

 xe x   e x dx

 xe x  e x  e x ( x  1)  C.


x

10.

2 x

e dx .

x 3 ln x  e dx.

8. 11.

x

n  x n 1

e

9.

dx .

x2  e x dx.

e

12.

2 x 1

x

2

dx .

 3x dx .

Të njesohen integralet 13.

e

15.

 x sin 2x dx.

2x

( x 2  3x  4) dx .

14.

 (x

16.

x

2

2

 x  1) sin x dx.

cos 5x dx .

Zgjidhja. x 2  3x  4  u 13.  e 2 x ( x 2  3x  4)dx  (2x  3)dx  du v   e 2 x dx  v 

 1 2x e 2

1 2x 2 e ( x  3x  4) 2




7

1 2x 1 1 e (2x  3)dx  e 2 x ( x 2  3x  4)  I1 .  2 2 2 2x  3  u

1 1 I1   e (2x  3)dx  2dx  du  e 2 x (2x  3)   2  e 2 x dx 2 2 1 v  e2x 2 2x



1 2x 1 1 e (2x  3)   e 2 x dx  e 2 x (2x  3)  e 2 x . 2 2 2

Përfundimisht, I  

1 2x 2 1 1 e ( x  3x  4)  e 2 x (2x  3)  e 2 x 2 4 4 1 1 2x e (2x 2  6x  8  2x  3  1)  e2 x ( x 2  2x  6)  C. 4 2 x2  x  1  u

14.

2  ( x  x  1) sin xdx 

(2x  1)dx  du v   sin xdx

  cos x ( x 2  x  1)

v   cos x   cos x (2x  1)dx  ( x 2  x  1) cos x  I1 ,

ku

I1   (2x  1) cos xdx 

2x  1  u 2dx  du v   cos xdx

 (2x  1) sin x  2  sin xdx

v  sin x  (2x  1) sin x  2 cos x .

Pra, I  ( x 2  x  1) cos x  (2x  1) sin x  2 cos x  (1  x 2  x ) cos x  (2x  1) sin x  C.


8

x u dx  du

15.

 x sin 2xdx  v   sin 2xdx v





1 cos 2x 2

x cos 2x 1  sin 2x  C. 2 4 x2  u

16.

2  x cos 5x dx  2xdx  du  1 v  sin 5x 5



x cos 2x 1   cos 2xdx 2 2

x 2 sin 5x 2   x sin 5x dx 5 5

x 2 sin 5x 2  I1 , ku 5 5

x u

1 1 I1   x sin 5xdx  dx  du   x cos 5x   cos 5xdx 5 5 1 v   cos 5x 5



1 1 x cos 5x  sin 5x. 5 25

Pra, I 

x 2 sin 5x 2  1 1     x cos 5x  sin 5x  5 5 5 25 



x 2 sin 5x 2 2  x cos 5x  sin 5x 5 25 125

 x2 2  2   x cos5x  C.  sin 5x  25  5 125 


9


x

15.

 (x

2

sin 3x dx. 2

 x  1) cos 2x dx .

14.

 ( x  1) cos( x  2) dx.

16.

x

x

( x 2  x  3)dx .

Të njehsohen integralet 17.

e

x

20.

e

x

sin x dx.

23.

e

x

cos2 x dx .

cos x dx .

18.

 sin(ln x ) dx.

19.

 cos(ln x ) dx.

21.

e

22.

e

x

cos x dx .

x

sin 2 x dx .

Zgjidhja. cos x  u

17.

e

x

cos x dx   sin xdx  du  e  x cos x   e  x sin x dx v  e  x

 e  x cos x  I1 ,

ku sin x  u I1   e

x

sin x dx  cos xdx  du  e  x sin x   e  x cos x dx v  e  x

 e  x sin x  I .

Pra, I  e  x cos x  ( e  x sin x  I ) I  e  x cos x  e  x sin x  I 2 I  e  x (sin x  cos x ) ex I  (sin x  cos x )  C. 2 18.  sin(ln x )dx. 19.  cos(ln x )dx . Në vijim, njëkohësisht do të zgjidhim integralet në detyrat 18, 19. Le të shënojmë: I1   sin(ln x )dx ; I 2   cos(ln x )dx. Zgjidhim I1 . Merret:


10

sin(ln x )  u cos(ln x ) cos(ln x ) I1   sin(ln x )dx  dx  du  x sin(ln x )   x dx x x vx  x sin(ln x )   cos(ln x )dx  x sin(ln x )  I 2 .

Pra, kemi: I1  x sin(ln x )  I 2 gjegjësisht I1  I 2  x sin(ln x )

(1)

Zgjidhim I 2 . Merret: cos(ln x )  u I 2   cos(ln x )dx 

 sin(ln x ) sin(ln x ) dx  du  x cos ln x   x dx x x vx

 x cos(ln x )   sin(ln x ) dx  x cos(ln x )  I1 .

Pra, I 2  x cos(ln x )  I1 , gjegjësisht,  I1  I 2  x cos(ln x ).

(2)

Nga (1), (2) merret sistemi:  I1  I 2  x sin(ln x )   I1  I 2  x cos(ln x )

me zgjidhjen e të cilit merret: x x I1  (sin(ln x )  cos(ln x )), I 2  (sin(ln x )  cos(ln x )). 2 2 x x 20.  e sin xdx. 21.  e cos xdx . Veprojmë ngjashëm si në detyrat paraprake. Shënojmë: I1   e x sin xdx, Zgjidhim integralin I1 .

I 2   e x cos xdx .


11

sin x  u I1   e sin xdx  cos xdx  du  e x sin x   e x cos xdx  e x sin x  I 2 . x

v  ex

Pra, I1  I 2  e x sin x .

(1)

Zgjidhim integralin I 2 . cos x  u I 2   e cos xdx   sin xdx  du  e x cos x   e x sin xdx x

v  ex

 e x cos x  I1 .

Pra  I1  I 2  e x cos x

(2)

Nga (1) dhe (2) merret sistemi:  I1  I 2  e x sin x  x  I1  I 2  e cos x

me zgjidhjen e të cilit merret: I1 

ex ex (sin x  cos x ), I 2  (sin x  cos x ). 2 2

Duke vepruar si më sipër tregoni se:

e

ax

sin bxdx 

a sin bx  b cos bx ax e a 2  b2

a cos bx  b sin bx ax e . a 2  b2 1  cos 2x 1 1 22.  e x sin 2 xdx   e x  dx   e x dx   e x cos 2xdx 2 2 2 1 x 1  cos 2x  2 sin 2x  x  e    e  C. 2 2 5  Shënim. Shprehja e fundit mund të transformohet në ex (sin 2 x  2 sin x cos x  2)  C. 5 1  cos 2x 1 1 23.  e x cos2 xdx   e x  dx   e x dx   e x cos 2xdx 2 2 2

e

ax

cos bxdx 


12



1 x 1  cos 2x  2 sin 2x  x e   e 2 2 5 



1 x cos2 x  sin 2 x  4 sin x cos x  e 1   2  5 



1 x e (cos2 x  2 sin x cos x  2)  C. 5


e

20.

 sin

23.

e

2x

x

cos 3x dx .

18.

e

2x

sin 3x dx.

19.

 cos (ln x ) dx.

2

21.

e

ax

sin 2 x dx .

22.

e

24.

e

x

(ln x ) dx.

sin 4 x dx.

2

ax

cos2 x dx .

cos4 x dx .

Të njehsohen integralet 25.

27.

 x  a dx.  arctan x dx.

30.

 x arc tgx dx.

33.

 (1  x

24.

2

2

dx 2 2

)

.

 x  a dx.  x arccos x dx.

26.

31.

2  x (arc tgx ) dx.

32.

34.

x

28.

2

2

2

2

29.

1  x 2 dx . 35.



 arcsin x dx.  (arcsin x ) dx. 2

x2  (1  x 2 )2 dx. x5 dx . 1  x4

Zgjidhja.

24. I 



x 2  a2  u xdx xdx x 2  a 2 dx   du  x  x 2  a 2   x  2 2 x a x 2  a2 vx

 x x 2  a2  

x 2  a2  a2 x 2  a2

dx

 x 2  a2 dx  x x 2  a2    dx  a 2   2 2 2 x a x  a2 

 x x 2  a 2   x 2  a 2 dx

  


dx

a 2 

2

x a

13

 x x 2  a 2  I  a 2 ln|a  x 2  a 2 |.

2

Pra 2 I  a x 2  a 2  a 2 ln| x  x 2  a 2 |

x x 2  a2 a2  ln| x  x 2  a 2 | C. 2 2 25. Duke vepruar ngjashëm si në detyrën paraprake merret: I 



x 2  a 2 dx 

x x 2  a2 a2  ln| x  x 2  a 2 | C. 2 2 arcsin x  u

26.

1

 arcsin xdx 

1  x2 vx

dx  du  x arcsin x  

xdx 1  x2

 x arcsin x  1  x 2  C. arctan x  u

27.

dx

 arctan xdx  1  x

2

 du  x arctan x  

xdx 1  x2

vx

 x arctan x 

1 ln(1  x 2 )  C. 2 arccos x  u, 

28. I   x arccos xdx  2



v

1 1  x2

dx  du

x3 3

x3 1 x3 x3 1 arccos x   dx  arccos x  I1 , ku 2 3 3 3 3 1x

I1   x 2

x2  u 2xdx  du

x 1x

2

dx 

v



xdx 1  x2

v   1  x2

  x 2 1  x 2   1  x 2  2dx


14

 x 2 1  x 2 

Pra I 

2 (1  x 2 )3 . 3

x3 1 2 arccos x    x 2 1  x 2  (1  x 2 )3 3 3 3

   C. 

(arcsin x )2  u

29.

 (arcsin x ) dx  2

2 arcsin x 1  x2 vx

 x (arcsin x )2  2 x

dx  du

arcsin x 1x

2

dx  x (arcsin x )2  2 I1 ,

ku I1   arcsin x 

x 1  x2

arcsin x  u, dx  v



  1  x 2  arcsin x   1  x 2 

x 1x dx

2

1x

dx 1  x2

 du

dx  v   1  x 2

2

  1  x 2  arcsin x  x.

Pra, I  x (arcsin x )2  2 1  x 2  arcsin x  2x  C.

arctan x  u dx x2 x2 dx  du   arctan x   30.  x arctan x dx  2  2 2 1  x2 1x x2 v 2 x2 1  1    arctan x   1  dx 2 2  1  x 2  x2 1 1 x2  1 1   arctan x  x  arctan x   arctan x  x  C. 2 2 2 2 2


15

(arctan x )2  u 2 arctan x dx  du 31. I   x (arctan x )2 dx  1  x2 x2 v 2 2 2 x x arctan x x2  (arctan x )2   dx  (arctan x )2  I1 , 2 2 1  x2 1 arctan x  u, dx  du 2 x 1  x2 ku I1   arctan x  dx  1  x2 x2 v dx  v  x  arctan x 1  x2 ( x  arctan x )  ( x  arctan x ) arctan x   dx 1  x2 1 x arctan x  x arctan x  (arctan x )2   dx   dx 2 2 1x 1  x2 1 (arctan x )2  x arctan x  (arctan x )2  ln(1  x 2 )  2 2 1 1  x arctan x  (arctan x )2  ln(1  x 2 ). 2 2 Pra x2 1 1 I  (arctan x )2  x arctan x  (arctan x )2  ln(1  x 2 ) 2 2 2 2 x 1 1  (arctan x )2  x arctan x  ln(1  x 2 )  C. 2 2 x u dx  du 2 x xdx 1 x 1 dx xdx 32.  dx   x   v    2 2 2 2 2 2 2 21x 2 1  x2 (1  x ) (1  x ) (1  x ) v

1 1  2 1  x2

1 x 1  arctan x  C. 2 21x 2 dx 1  x2  x2 1  x2 x2  dx  dx  33. I    (1  x 2 )2  (1  x 2 )2 dx (1  x 2 )2  (1  x 2 )2 


16

dx x2   1  x 2  (1  x 2 )2 dx. x2 dx e zgjidhëm në detyrën paraprake, prandaj Integrali  (1  x 2 )2 1 x 1 arctan x 1 x I  arctan x   arctan x    C. 2 21x 2 2 2 1  x2 x u 

dx  du

34. I   x 2 1  x 2 dx   x  x  1  x 2 dx  v   x 1  x 2 dx v

(1  x 2 ) 1  x 2 3

x 1 (1  x 2 ) 1  x 2   (1  x 2 ) 1  x 2 dx 3 3 x 1 1  (1  x 2 ) 1  x 2   (1  x 2 )dx   x 2 1  x 2 dx 3 3 3 x 1 1  (1  x 2 ) 1  x 2  I1  I , 3 3 3 

ku

I1 

Pra



1  x 2 dx 

x 1 1  x 2  ln| x  1  x 2 |. 2 2

1 x 1x 1 I  (1  x 2 ) 1  x 2  1  x 2  ln| x  1  x 2 | 3 3 32 6 x (1  2x 2 ) 1 I  1  x 2  ln| x  1  x 2 | C. 8 8 x2  u 2xdx  du 5 3 x x x3 35. I   dx   x 2  dx  v   1  x 4 dx 1  x4 1  x4 1 v 1  x4 2 x2 1 x2  1  x 4   2 x 1  x 4 dx   1  x 4  I1 . 2 2 2 I


17

x2  t

1 2 dt   1  t dt 2 xdx  2 1 1 t x2  1   arcsin t  1  t 2   arcsin x 2  1  x4 . 22 2 4  4 Pra, x2 1 x2 I  1  x 4  arcsin x 2  1  x4 2 4 4 x2 arcsin x 2  1  x4   C. 4 4 I1   x 1  x dx   x 1  ( x ) dx  4

2 2

Detyra për ushtrime Të njehsohen integralet: x arcsin x

25.



27.

x 2ex  ( x  2)2 dx.

29.

 (a

1  x2

2

dx .

dx .  x 2 )3

26.  arcsin x  arccos xdx. 28.

x 2 arctan x  1  x 2 dx.

30.

x

2

a 2  x 2 dx .

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Analiza Matematike 2 Integrali i Pacaktuar a.sh .Shabani

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