How to integrate $\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$

Question

I am asked to solve the following integral:

$$\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$$

I've made a couple of guesses but none of them took me anywhere. One of my ideas was to do, at first, an u-substitution:

$$ u = 1+x^2\\ du = 2x dx\\ \frac{du}{2} = x dx\\ \\ x = 0 \Rightarrow u = 1\\ x = a \Rightarrow u = 1+a^2 $$

but then there's a $\arctan(x)$ left there untouched. Does anyone have an idea I can pursue?

Source: James Stewart's Calculus book.

Thank you.

Answer 1

Set $u = \arctan(x)$, $du = \frac{dx}{1+x^2}.$ Your integral becomes $$\int_0^{\pi/2}\frac{u\tan(u)}{1+\tan^2(u)}du = \int_0^{\pi/2}u\tan(u)\cos^2(u)du = \int_0^{\pi/2}u\sin(u)\cos(u)du \\= \frac{1}{2}\int_0^{\pi/2}u\sin(2u)du=\frac{1}{2}\left[-\frac{1}{2}u\cos(2u)+\frac{1}{4}\sin(2u)\right]_0^{\pi/2}=\frac{\pi}{8}.$$

Answer 2

Let us integrate by parts and put

$u'=\frac{x}{(1+x^2)^2}$

and

$v=arctan(x).$

thus

the integral becomes

$$I=[-\frac{1}{2(1+x^2)}arctan(x)]_0^\infty$$

$$+\frac{1}{2}\int_0^\infty\frac{dx}{(1+x^2)^2}$$

$$=\frac{1}{2}\int_0^\infty\frac{1+x^2-x^2}{(1+x^2)^2}dx$$

$$=\frac{\pi}{4}+\frac{1}{4}\int_0^\infty x \frac{-2x}{(1+x^2)^2}dx$$

$$=\frac{\pi}{4}+\frac{1}{4}([\frac{x}{1+x^2}]_0^\infty-\int_0^\infty\frac{1}{1+x^2}dx)$$

$$\color{red}{=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8}}$$

Answer 3

With the substitution $x=\frac{1}{t}$ we get $$ I=\int_{0}^{+\infty}\frac{x\arctan(x)}{(1+x^2)^2}\,dx = \int_{0}^{+\infty}\frac{t\left(\frac{\pi}{2}-\arctan(t)\right)}{(1+t^2)^2}\,dt $$ hence by symmetry $$ I = \frac{\pi}{8}\int_{0}^{+\infty}\frac{2u}{(1+u^2)^2}\,du = \color{red}{\frac{\pi}{8}}.$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{x\arctan\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x & \,\,\,\stackrel{x\ \equiv\ \tan\pars{\theta}}{=}\,\,\, {1 \over 2}\int_{0}^{\pi \over 2}\theta\sin\pars{2\theta}\,\dd\theta \\[5mm] & = {1 \over 4}\int_{0}^{\pi \over 2} \bracks{\theta\sin\pars{2\theta} + \pars{{\pi \over 2} - \theta}\sin\pars{2\bracks{{\pi \over 2} - \theta}}} \,\dd\theta \\[5mm] & = {1 \over 8}\,\pi\int_{0}^{\pi \over 2}\sin\pars{2\theta}\,\dd\theta = \bbx{\ds{{1 \over 8}\,\pi}} \end{align}