Integration by parts

**saiyanmx89** · February 22nd 2009, 01:14 PM

$\int \frac {x^3e^{x^2}}{(x^2+1)^2}dx$

I did this:
u= $\frac {1}{(x^2+1)^2}$ -> du= $\frac {-4x}{(x^2+1)^3}$
dv= $x^3e^{x^2}$ -> v= $\frac {1}{2}(x^2-1)e^{x^2}$

Am I going in the right direction?? This is so frustrating...

**o_O** · February 22nd 2009, 01:33 PM

Try:
$u = x^2e^{x^2} \ \Rightarrow \ du = \left(2xe^{x^2} + 2x^3e^{x^2}\right)dx = 2xe^{x^2} (1 + x^2)dx$
$dv = \frac{x}{(1+x^2)^2} \ dx \ \Leftrightarrow \ v = -\frac{1}{2(x^2+1)}$

So: $\int \frac{x^3e^{x^2}}{(x^2+1)^2} \ dx = -\frac{x^2e^{x^2}}{2(x^2+1)} + \int xe^{x^2} dx$

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