# How to find the integral of x^3 e^(x^2) / (x^2 + 1)^2 ?

1. Jun 7, 2008

### rocomath

Ok wow, I can't solve this :(

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

I have let u equal almost everything possible.

$$u=x^2$$
$$d=2xdx$$

$$u=e^{x^2}$$
$$\ln u=x^2$$
$$\frac{du}{u}=2xdx$$

$$u=x^2+1 \leftrightarrow u-1=x^2$$
$$\ln u=\ln{(x^2+1)}$$
$$\frac{du}{u}=\frac{2xdx}{x^2+1}$$

2. Jun 7, 2008

### Acetv

rocomath, is the integral given as a definite integral? It is possible to find the value of the definite integral by using integration by parts (think about the change of variables a bit). I don't really see an easy u-substitution that's going to help you out, but maybe I'm just blind.

3. Jun 7, 2008

### rocomath

OH, lol. I forgot to include what dV was ...

But yes, I've been trying parts on it for a while now ... still nothing!

4. Jun 7, 2008

### tiny-tim

Hi rocomath!

Go with $$u=x^2$$ again …

you should get $$\int\frac{ue^u}{(1 + u)^2}du$$

Then use partial fractions, and try the obvious.

5. Jun 7, 2008

### rocomath

Uh!!! I didn't even think of using Partial Fractions bc the people in the group said they weren't doing that yet ... so frustrated, lol.

tiny-tim you're a life saver!

6. Jun 7, 2008

### rocomath

Ok ... so I have

$$\frac 1 2\int\frac{te^t}{(t+1)^2}dt$$

Now using Partial Fractions and leaving the constant 1/2 out for now:

$$\frac{te^t}{(t+1)^2}=\frac{A}{t+1}+\frac{B}{(t+1)^2}$$

$$te^t=A(t+1)+B$$

if t = -1

$$B=-\frac 1 e$$

I'm having a hard time solving for A ...

I know what B is, so if t = 0

$$A=-\frac 1 e$$ ???

... hmm, definitely not right!

I've never done a problem that had a product on the left side.

Last edited: Jun 7, 2008
7. Jun 7, 2008

### Acetv

The method of partial fractions won't work like it normally does here since you can't find integers $A$ and $B$ such that

$$t e^t = A (t + 1) + B$$

You can, however, solve for $A$ and $B$ in terms of your variable $t$, but it doesn't really make the problem any easier. We might find

$$B = - e^{-1}$$

And so plugging that in to the equation above we find

$$A = \frac{t e^t - e^{-1}}{t + 1}$$

I still suggest trying the method of integration by parts. Hint: can you get the entire denominator $(x^2+1)^2$ into $dv$ such that you can easily integrate it?

8. Jun 7, 2008

### rocomath

Ok, got it!

Uh, sucha killer. Thanks for the hint Acetv, whewww!!!

9. Jun 8, 2008

### Gib Z

We still could have used ordinary partial fractions - I'm not sure why you thought it was even possible to find constants A and B for te^t = A(t+1) + B, but tiny-tims suggestion was to use partial fractions on what you've been taught to use it on - Rational functions.

IE Only the $$\frac{t}{(t+1)^2}$$. After that, multiply each term by e^t.

You should have got $$\frac{e^t}{t+1} - \frac{e^{t}}{(t+1)^2} = \frac{1}{e} \left( \frac{e^u}{u} - \frac{e^u}{u^2} \right)$$ where u = t+1, which are simple to integrate.