# Integral: Hint

1. Dec 26, 2007

### rocomath

Integral: Hint plz

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

$$\int\frac{\tan^{ - 1}x}{(1 + x^2)}\frac{x}{(1 + x^2)^2}dx\left\{\begin{array}{cc}x = \tan{r} \leftrightarrow r = \tan^{ - 1}x \\ dr = \frac {dx}{1 + x^2}\end{array}\right\}$$

Let's not even get into what I did next and how much work I put into this Integral. This is definitely not the right way to go, hmm ...

Ok I think I see it now ... omg.

Last edited: Dec 26, 2007
2. Dec 26, 2007

### arildno

Well, therefore, you have:
$$\int\frac{x\tan^{-1}(x)}{(1+x^{2})^{2}}\frac{dx}{1+x^{2}}=\int\frac{\tan(r)r}{(1+\tan^{2}(r))^{2}}dr=\int\cos^{4}(r)\tan(r)rdr=\int{r}\cos^{3}(r)\sin(r)dr=\frac{1}{4}\int\cos^{4}(r)dr-\frac{r}{4}\cos^{4}(r)$$

Agreed?

3. Dec 26, 2007

### rocomath

On my previous try, after the 2nd to last step of yours, I did it differently; I would like to try your way, but I'm not sure what you did.

I did ... $$\int r\sin{r}(1-\sin^{2}r)\cos{r}dr$$

Nvm, I see what you did. UHHH!!! I can't believe I spent more than an hour on this problem. Thanks arildno :-]

Last edited: Dec 26, 2007
4. Dec 28, 2007

### unplebeian

Hi arildno,

I followed all but how you went from the second last step to the last step, as there is an 'r' in the integrand. Are you using some sort of gamma function there. That's the closest I've seen to int( sin^m(x) x cos^n(x) dx)

Thanks.

5. Dec 28, 2007

### arildno

I'm just using the product rule:
$$u(r)=r, v'(r)=\cos^{3}(r)\sin(r)\to{v}(x)=-\frac{1}{4}\cos^{4}(r)$$

6. Dec 28, 2007

### unplebeian

You're a freakin genius. I totally didn't see that. I was trying to solve it using a gamma function. I think there should be something related to a gamma function too as you still have to solve the int(cos^4(r) dr) which is part of the question.

Thanks for the help.