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

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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 x d x Let ; then . 2 1 u x = 2 du xdx = 7 7 8 1 2 (2 1) 8 x x dx u du u = = + C 8 1 (2 1) 8 x C = + 3. 2 2 1 3 x dx x x + + + Let ; then . 2 3 u x x = + + ( ) 2 1 du x dx = + 1 2 2 2 1 1 2 3 x dx du u C u x x + = = + + + 2 2 3 x x C = + + + 4. ( ) ( ) 6 2 3 1 x x x + + + dx Let ; then 2 3 u x x = + + 1 (2 2) ( 1) 2 du x dx du x dx = + = + . ( ) ( ) 6 2 6 1 1 3 1 2 14 7 x x x dx u du u + + + = = + C ( ) 7 2 1 2 3 14 x x C = + + + 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 u xe dx e du = − = − e u + C = − e x 2 + C 7. 2 4 x x dx 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 ) 3/2 + C 8. ( ) 3 1 ln x dx x + Let 1 ln u x = + ; then du = 1 x dx . ( ) 3 3 4 1 ln 1 4 x dx u du u C x + = = + 4 1 (1 ln ) 4 x C = + + 9. 1 2 1 dx x + Let 2 1 u x = + ; then 1 2 2 du dx du dx = = . 1/ 2 1 1 1 2 2 2 1 du dx u du x u = = + 1/ 2 1/ 2 1 (2 ) 2 u C u C = + = + 2 1 x C = + + 10. ( x 3 6 x ) 7 ( x 2 2) dx Let u = x 3 6 x ; then 2 2 1 (3 6) ( 2) 3 du x dx du x dx = = . ( x 3 6 x ) 7 ( x 2 2) dx = 1 3 u 7 du = 1 24 u 8 + C = 1 24 ( x 3 6 x ) 8 + C 11. 2 x xe dx Let 2 u x = ; then 1 2 2 du xdx du xdx = = . 2 2 1 1 2 2 2 u x u x e xe dx e du C e C = = + = +

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