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

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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