# 4 du dx so 12 32 32 1 4 1 4 1 2 4 3 4 1 6 x dx u du u

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4 du dx = So 1/2 3/2 3/2 1 4 1 4 1 2 4 3 (4 1) . 6 x dx u du u C x C = = = 6. 1 3 5 I dx x = + Let 3 5 u x = then 1 3 , 3 du dx dx du = and 1 3 5 1 1 3 1 ln 3 1 ln 3 5 3 I dx x du u u C x C = + = = + = + Check: 1 3 1 3 3 5 3 5 I x x ′ = + + 7. Let u = 1 x . Then du = dx or dx = du . So 1 1 . x u x e dx e du e C = − = − + 8. 5 2 ( 1) 3( 1) 5 I x x dx = + + Let 1 u x = then du dx = and 5 2 6 3 6 3 ( 3 5) 5 6 ( 1) ( 1) 5( 1) 6 I u u du u u u C x x x C = + + = + + + = + + + Check: 5 2 1 (6)( 1) 3( 1) 5 6 I x x ′ = + 9. Let 2 . u x = Then 2 du x dx = or 1 . 2 du xdx = 2 2 2 1 2 1 2 1 . 2 x x u u x xe dx e xdx e du e du e C = = = = 10. 2 1 2 x I xe dx = Let 2 1 u x = then 2 du xdx = and 2 2 1 1 (2 ) x u u x I e x dx e du e C e C = = = + = + Check: ( ) ( ) 2 1 2 x I e x ′ =
Chapter 5. Integration 495 11. Let 2 1. u t = + Then 2 du t dt = or 1 . 2 du t dt = 2 5 2 5 5 5 2 6 ( 1) ( 1) 1 2 1 2 ( 1) . 12 t t dt t t dt u du u du t C + = + = = + = 12. 2 3 8 I t t dt = Let 2 8 u t = + then 2 du t dt = and 2 3/ 2 2 3/ 2 3 8 (2 ) 2 3 2 ( 8) I t t dt u du u C t C = + = = + = + + Check: 2 1 2 2 1 2 3 ( 8) (2 ) 3 ( 8) 2 I t t t t ′ = + = + 13. Let 3 1. u x = + Then 2 3 du x dx = or 2 1 . 3 du x dx = 2 3 3/4 3 3/4 2 3/4 3/4 3 7/4 ( 1) ( 1) 1 3 1 3 4( 1) 21 x x dx x x dx u du u du x C + = + = = + = 14. 6 5 1 x I x e dx = Let 6 1 u x = then 5 6 du x dx = − and 6 6 6 5 1 1 5 1 1 ( 6 ) 6 1 6 1 6 1 6 x x u u x I x e dx e x dx e du e C e C = = − = − = − + = − + Check: ( ) 6 6 1 5 5 1 1 6 6 x x I e x x e ′ = − = 15. Let 5 1. u y = + Then 4 5 , du y dy = or 4 1 . 5 du y dy = 4 4 5 5 5 2 1 2 1 1 1 1 2 5 2 1 5 2 ln 1 5 y dy y dy y y du u du u y C = + + = = = + + 16. ( ) 2 2 3 5 y I dy y = + Let 3 5 u y = then 2 3 du y dy = and ( ) ( ) 2 2 3 2 2 3 2 1 3 5 1 1 (3 ) 3 5 1 3 1 3 1 3( 5) y I dy y y dy y u du u C C y = + = + = = − + = − + +
496 Chapter 5. Integration Check: ( ) ( ) ( ) ( ) 2 3 2 2 2 3 1 1 5 3 3 5 I y y y y ′ = + = + 17. Let 2 2 5. u x x = + + Then 2 2 2( 1), du x x dx = + = + or 1 ( 1) . 2 du x dx = + 2 12 2 12 12 12 2 13 ( 1)( 2 5) ( 2 5) ( 1) 1 2 1 2 ( 2 5) 26 x x x dx x x x dx u du u du x x C + + + = + + + = = + + = 18. ( ) 3 2 3 1 x x I x e dx = Let 3 u x x = then 2 (3 1) du x dx = and ( ) ( ) 3 3 3 2 2 3 1 3 1 x x x x u u x x I x e dx e x dx e du e C e C = = = = + = + Check: ( ) ( ) 3 2 3 1 x x I e x ′ = 19. Let 5 4 5 10 12. u x x x = + + + Then 4 3 4 3 5 20 10 5( 4 2), du x x dx x x = + + = + + or 4 3 5 4 4 3 5 4 4 3 5 4 5 4 3 12 6 5 10 12 3( 4 2) 5 10 12 1 3 ( 4 2) 5 10 12 1 1 3 5 3 1 5 3 ln 5 10 12 5 x x dx x x x x x dx x x x x x dx x x x du u du u x x x C + + + + + + + = + + + = + + + = = = + + + + 20. 3 4 2 10 5 6 x x I dx x x = + Let 4 2 6 u x x = + then 3 (4 2 ) du x x dx = and 3 4 2 3 4 2 1/ 2 1 2 4 2 10 5 6 (5/ 2)(4 2 ) 6 5 1 2 5 2 5 2 1 2 5 5 6 x x I dx x x x x dx x x du u u du u C u C x x C = + = + = = = + = + = + + Check: ( ) ( ) ( ) ( ) 1 2 4 2 3 3 4 2 1 5 6 4 2 2 5 2 6 I x x x x x x x x ′ = + = +
Chapter 5. Integration 497 21. Let 2 2 6. t u u = + Then 2 2 2( 1), dt u u du = = or 1 ( 1) .