Integration by Parts: Solving \int \frac{x^3e^{x^2}}{(x^2+1)^2}

[Zhalfirin88]

Homework Statement

[tex] \int \frac{x^3e^{x^2}}{(x^2+1)^2} [/tex]

The Attempt at a Solution

Well, this problem is hard, so I thought to use u = x³e^x2
so du = x²e^x2(3+2x²) dx
and dv = (x²+1)^-2 then v = -2(x²+1)^-1 Please check v though to make sure my algebra is right.

so then using the by parts formula:
[tex] \int \frac{x^3e^{x^2}}{x^2+1} = \frac{x^3e^{2x}}{2(x^2+1)} - \int \frac{x^2e^{x^2}(3+2x^2)}{2(x^2 +1)} [/tex] But where do I go from here?

 

[Mark44]

I wouldl start with u = x²/(x² + 1)² and dv = xe^x2. No guarantee that this will get you anywhere, but a useful strategy is to make dv the most complicated expression that you can integrate.

 

[l'Hôpital]

Notice the fact that you have x^3 on the top. Try a u-sub with u = (1 + x^2). After that, use parts cleverly and you'll get your answer.

 

[Dick]

This is one of those cases where there are several plausible alternatives for picking the parts, but only one works. You just have to mess around. Hint: if you pick u=x^2*e^(x^2) then du=(2xe^(x^2)+2x^3e^(x^2))dx=2(1+x^2)*e^(x^2)*x*dx. Notice the (1+x^2) factor in du. You need that.

 

[LCKurtz]

and dv = (x²+1)^-2 then v = -2(x²+1)^-1 Please check v though to make sure my algebra is right.

You don't get that v for your choice of dv. You probably meant

[tex]\frac{(x^2+1)^{-1}}{-1}[/tex]

but that's still wrong because you would need a 2x in the in the numerator of dv.