sol9 - Chapter 9 Exercises 9.1 1. 7. 2 Let u = x + 4 ; then...

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273 Chapter 9 Exercises 9.1 1. Let u 2 x ( x 2 + 4) 5 dx = x 2 + 4; then du = 2 x dx . 2 x ( x 2 + 4) 5 dx = u 5 du = 1 6 u 6 + C = 1 6 ( x 2 + 4) 6 + C 2. 7 2( 2 1 ) x xd x Let ; then . 21 ux =− 2 du xdx = 77 8 1 2 1 ) 8 x xu d uu −= =+ ∫∫ C 8 1 (2 1) 8 x C + 3. 2 3 x dx xx + ++ Let ; then . 2 3 ux x =++ () du x dx 1 2 2 1 2 3 x dx du u C u + == + 2 23 x xC + + 4. 6 2 31 x + d x Let ; then 2 3 1 2) ( 2 du x dx du x dx =+ ⇒ =+ . 6 26 11 4 7 x d d u u + = = + C 7 2 1 14 x + + 5. Let u 3 x 2 e ( x 3 1) dx = x 3 1; then du = 3 x 2 dx . 3 x 2 e ( x 3 1) dx = e u du = e u + C = e x 3 1 + C 6. Let u 2 xe x 2 dx x 2 ; then 2 2 du xdx du xdx ⇒− = 2 2 x ed x e d u e u + C e x 2 + C 7. 2 4 x x Let u = 4 x 2 ; then 1 2 2 du xdx du xdx = . x 4 x 2 dx 1 2 u 1/2 du 1 3 u 3/2 + C 1 3 (4 x 2 ) + C 8. 3 1l n x dx x + Let n = + ; then du = 1 x dx . 3 34 n 1 4 x dx u du u C x + = 4 1 (1 ln ) 4 x C = 9. 1 dx x + Let = + ; then 1 2 2 du dx du dx =⇒ = . 111 22 du dx u du + 1 ) 2 uC u ⎛⎞ C = += + ⎜⎟ ⎝⎠ x C = 10. ( x 3 6 x ) 7 ( x 2 2) dx Let u = x 3 6 x ; then 1 (3 6) ( 2) 3 du x dx du x dx =− ⇒=− . ( x 3 6 x ) 7 ( x 2 dx = 1 3 u 7 du = 1 24 u 8 + C = 1 24 ( x 3 6 x ) 8 + C 11. 2 x x x Let 2 = ; then 1 2 2 du xdx du xdx =⇒= . 2 u x e x x u C e C = =+= +
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Chapter 9: Techniques of Integration ISM: Calculus & Its Applications, 11e 274 12. x e dx x Let u = x ; then 11 2 2 du dx du dx xx =⇒ = . 22 x uu e dx e du e C x == ∫∫ + 2 x eC =+ 13. ln(2 x ) x dx Let u = ln (2 x ); then du = 1 2 x (2) dx = 1 x dx . ln(2 x ) x dx = u du = u 2 2 + C = (ln(2 x )) 2 2 + C 14. ln x dx x Let u = ln x ; du = 1 x dx . 1/2 3/2 ln 2 3 x dx u du u C x + = 2 3 (ln x ) + C 15. x 4 x 5 + 1 dx Let then 5 1; ux 44 1 5 5 du x dx du x dx =⇒= x 4 x 5 + 1 dx = 1 5 du u = 1 5 ln u + C = 1 5 ln x 5 + 1 + C 16. 2 1 x dx x + Let u = x 2 + 1; then 1 2 2 du xdx du xdx . 2 2 1 x dx u du u C x ⎛⎞ ⎜⎟ ⎝⎠ + + C 2 (1 ) uCx =++ 17. x 3 6 x + x 2 ) 2 dx Let u = 1 6 x + x 2 ; then 1 (6 2) ( 3 ) 2 du x dx du x dx =−+ = . 2 31 1 1 1 6 ) x dx du C u u = =− + −+ 1 2(1 6 x + x 2 ) + C 1 2 12 x + 2 x 2 + C 18. 15 2 (2 ) x dx x + Let 1 2; = + then du x dx du x dx −− ⇒− = 6 5 2 ) 6 xu dx u du C x + = −= + 1 6 ( x 1 + 2) 6 + C 19. ln x dx x Let u = ln x ; then du = 1 x dx . () 1 2 ln ln ln( ) x dx dx dx xx x ∫∫ ∫ = 1 2 ln x x dx = 1 2 u du = 1 2 1 2 u 2 + C (ln ) or (ln ) uC x C 2 = += + + 20. x 2 3 x 3 dx Let 3 3 = ; then 1 3 3 du x dx du x dx = x 2 3 x 3 dx 1 3 du u 1 3 ln u + C 1 3 ln3 x 3 + C
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ISM: Calculus & Its Applications, 11e Chapter 9: Techniques of Integration 275 21. x 2 2 x x 3 3 x 2 + 1 dx Let u then = x 3 3 x 2 + 1; 2 (3 6 ) du x x dx =− 2 1 (2 ) 3 du x x dx ⇒=− 2 32 21 3 31 x xd dx u xx = −+ ∫∫ u 11 ln ln 3 1 33 uC x x C =+ + + 22. ln(3 x ) 3 x dx . Let ; then ln(3 ) u = x (3) 3 du dx dx x x == du dx x ⇒= 2 ln(3 ) 1 1 1 333 2 x dx udu u C x ⎛⎞ ⎜⎟ ⎝⎠ + 2 1 (ln3 ) 6 x C 23.
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This note was uploaded on 04/24/2010 for the course MATH 9374 taught by Professor Smith during the Spring '10 term at University of California, Berkeley.

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sol9 - Chapter 9 Exercises 9.1 1. 7. 2 Let u = x + 4 ; then...

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