Question

How do you integrate `∫(x^3ex^2)/(x^2 +1)^2dx`?

`∫(x^3e^(x^2))/(x^2 +1)^2dx` ?

Answer 1

`int(x^3e^(x^2))/(x^2+1)^2dx=e^(x^2)/(2(x^2+1))+c`

Explanation:

Here,

`I=int(x^3e^(x^2))/(x^2+1)^2dx`

`=int(x^2e^(x^2))/(x^2+1)^2*xdx`

Subst . `color(violet)(x^2=u=>2xdx=du=>xdx=1/2du`

So,

`I=int(ue^u)/(u+1)^2*1/2du`

`=1/2int((u+1-1)e^u)/(u+1)^2du`

`=1/2int{((u+1))/(u+1)^2-1/(u+1)^2}e^udu`

#=1/2{color(blue)(int1/(u+1)e^udu)-color(red) (int1/(u+1)^2e^udu)}#

Using Integration by parts: in the first integral

`I`=#1/2{color(blue)([1/(u+1)inte^u-int(-1)/(u+1)^2e^udu])- color(red)(int1/(u+1)^2e^udu)}#

`I`=#1/2{color(brown)(1/(u+1)e^u+color(blue) (int(1)/(u+1)^2e^udu)-color(red)(int1/(u+1)^2e^udu)}#

`I=1/2 *color(brown)(1/(u+1)e^u)+c`

Subst. back `color(violet)( u=x^2` we get

`I=1/2(1/(x^2+1))e^(x^2)+c`

Hence ,

`I=e^(x^2)/(2(x^2+1))+c`
.....................................................................................

Note: Change in the question is as below.

`x^3ex^2tox^3e^(x^2)`

Answer 2

The answer is `=(e^(x^2))/(2(x^2+1))+C`

Explanation:

Perform the substitution

`u=x^2`, `=>`, `du=2xdx`

The integral is

`I=int(x^3e^(x^2)dx)/(x^2+1)^2`

`=1/2int(ue^udu)/(u+1)^2`

Perform an integration by parts

`intwv'dx=wv-intw'vdx`

`w=ue^u`, `=>`, `w'=ue^u +e^u=e^u(u+1)`

`v'=1/(u+1)^2`, `=>`, `v=-1/(u+1)`

Therefore,

`I=-1/2(ue^u)/(u+1)+1/2inte^udu`

`=e^u/2-(ue^u)/(2(u+1))`

`=e^(x^2)/2-(x^2e^(x^2))/(2(x^2+1))+C`

`=1/2e^(x^2)((x^2+1-x^2))/(x^2+1)+C`

`=(e^(x^2))/(2(x^2+1))+C`