Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 57

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
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