# Integral: Hint

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.

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## Answers and Replies

arildno
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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?

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 :-]

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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?

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.

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

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

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.