`int x^5e^(x^2) dx` Find the indefinite integral by using substitution followed by integration by parts.

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marizi eNotes educator| Certified Educator

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 * 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`


`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`