Student Question

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

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

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