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