To evaluate the given integral problem `int x^5e^((x^2))dx` using u-substituion, we may let:
`u = x^2` then `du = 2x dx` or `(du)/2 = x dx`
Note that `x^5 = x^2*x^2*x` or ` (x^2)^2 *x` then
`x^5dx = (x^2)^2 * x dx`
Then, the integral becomes:
`int x^5e^((x^2))dx =int (x^2)^2...
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* e^((x^2)) * xdx`
Plug-in `u = x^2` then `du = 2x dx` , we get:
`int (x^2)^2 * e^((x^2)) * xdx =int (u)^2 * e^(u) * (du)/2`
Apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .
`int (u)^2 * e^(u) * (du)/2= 1/2int (u)^2 * e^(u) du`
Apply formula for integration by parts: `int f*g'=f*g - int g*f'` .
Let: `f =u^2` then `f' =2udu`
`g' =e^u du` then `g=e^u`
Applying the formula for integration by parts, we get:
`1/2int (u)^2 * e^(u) du =1/2*[ u^2 *e^u - int e^u * 2u du]`
`=1/2*[ u^2 e^u - 2 int e^u *u du]`
` = ( u^2 e^u )/2- 2/2 int e^u *u du`
` = ( u^2 e^u )/2- int e^u *u du`
Apply another set of integration by parts on `int e^u *u du` by letting:
`f =u` then `f’ = du`
`g’ = e^u du` then `g = e^u`
Then,
`int e^u *u du = u*e^u - int e^u du`
`= ue^u - e^u+C`
Applying `int e^u *u du =ue^u - e^u+C` , we get:
`1/2int (u)^2 * e^(u) du =( u^2 e^u )/2- int e^u *u du`
`=( u^2 e^u )/2-[ue^u - e^u] +C`
` =( u^2 e^u )/2-ue^u + e^u +C`
Plug-in `u = x^2` on `( u^2 e^u )/2-ue^u + e^u +C` , we get the complete indefinite integral as:
`int x^5e^((x^2))dx =((x^2)^2 e^((x^2)) )/2-x^2e^((x^2)) + e^((x^2)) +C`
`= (x^4 e^(x^2) )/2-x^2e^(x^2) + e^(x^2) +C`