# `int x^3e^(x^2)/(x^2+1)^2 dx` Find the indefinite integral

## Expert Answers

Given to solve,

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

let `t = x^2 => dt = 2x dx`

so,

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

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

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

`=int t*e^(t)/(t+1)^2 (1/2)dt`

let `u = t e^t => u'= e^t + te^t`

and `v'=(1/(t+1)^2)`

=> `v' = (t+1)^(-2) `

so `v=(t+1)^(-2+1) /(-2+1) = (t+1)^(-1) /(-1) `

=> `v= (-1)/(t+1)`

so , applying  integraion by parts we get ,

`int uv' = uv - int u'v `

so ,

`int t*e^(t)/(t+1)^2 (1/2)dt`

= `(1/2)[(t e^t )((-1)/(t+1)) - int (e^t + te^t)((-1)/(t+1)) dt]`

= `(1/2)[(-(t e^t )/(t+1)) + int (e^t + te^t)((1)/(t+1)) dt]`

=`(1/2)[(-(t e^t )/(t+1)) + int (e^t)(1 + t)((1)/(t+1)) dt]`

=`(1/2)[(-(t e^t )/(t+1)) + int (e^t) dt]`

=`(1/2)[(-(t e^t )/(t+1)) + (e^t)] +c`

but ` t = x^2`

so,

`1/2[(-(t e^t )/(t+1)) + (e^t)] +c`

`=1/2[((-x^2 e^(x^2) )/(x^2+1)) + (e^(x^2))] +c`

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

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

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