1. ## Improper Integral Halp!

I can't figure this one out: Integral from 0 to infinity of xarctanxdx/(1+x^2)^2

I know about lim(T--> Infinity) etc. etc., but I can't figure out the actual integral itself. It appears to be a d/dx arctanx = 1/1+x^2 situation, but I can't seem to get there. I can do it very long and slowly through Integration by parts..

1) u = x^2, du = 2x.

=1/2 (1/(1+u)^2)

and then integrate by parts, but it takes forever and seems wrong.

Any solutions?

2. Originally Posted by Sprintz
I can't figure this one out: Integral from 0 to infinity of xarctanxdx/(1+x^2)^2

I know about lim(T--> Infinity) etc. etc., but I can't figure out the actual integral itself. It appears to be a d/dx arctanx = 1/1+x^2 situation, but I can't seem to get there. I can do it very long and slowly through Integration by parts..

1) u = x^2, du = 2x.

=1/2 (1/(1+u)^2)

and then integrate by parts, but it takes forever and seems wrong.

Any solutions?
ok

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

let

$u = \tan ^{-1} x$

$\tan u = x \Rightarrow \sec ^2 u du = dx$ but $\sec u = \sqrt{x^2 +1 }$ so

$\int \frac{(u)\tan u \sec ^2 u }{\sec ^4 u } du$

$\int \frac{(u) \tan u }{\sec ^2 u} du$

by parts dv = tan u / sec^2 u , w =u so dw =du

$v =\int \frac{\tan u }{\sec ^2 u } du$

let s = sec u --------> ds = sec u tan u du so

$v = \int \frac{1}{s^3} ds \Rightarrow v = -\frac{s^{-2}}{2} \Rightarrow v = -\frac{\cos ^2 u}{2}$ ok

$\int \frac{u \tan u }{\sec ^2 u } du$ by parts as I said

so

$-\frac{u\cos ^2 u}{2} + \int \frac{\cos ^2 u}{2} du \Rightarrow -\frac{u\cos ^2 u}{2} +\frac{\sin 2u }{8} + \frac{u}{4}$

but $u = \tan ^{-1} x$

so $\int \frac{x \tan ^{-1} x }{(x^2+1)^2) } dx = -\frac{\tan ^{-1}x (\cos ^2(\tan ^{-1}x))}{2}+\frac{\sin 2(\tan ^{-1}x)}{8}+ \frac{\tan ^{-1} x}{4}$

the rest for you I find the integral

is there anything not clear ?