# Integration by parts

1. Jan 28, 2010

### Zhalfirin88

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution
Well, this problem is hard, so I thought to use u = x3ex2
so du = x2ex2(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:
$$\int \frac{x^3e^{x^2}}{x^2+1} = \frac{x^3e^{2x}}{2(x^2+1)} - \int \frac{x^2e^{x^2}(3+2x^2)}{2(x^2 +1)}$$ But where do I go from here?

2. Jan 28, 2010

### Staff: Mentor

I wouldl start with u = x2/(x2 + 1)2 and dv = xex2. No guarantee that this will get you anywhere, but a useful strategy is to make dv the most complicated expression that you can integrate.

3. Jan 28, 2010

### 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.

4. Jan 28, 2010

### 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.

5. Jan 28, 2010

### LCKurtz

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

$$\frac{(x^2+1)^{-1}}{-1}$$

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