# How to prove $\int_{0}^{1}{ dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}?$

Variation of Ahmed's integral

$$\int_{0}^{1}{\mathrm dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}\tag1$$

Making an attempt:

I can't think of what to do. It is too complicate.

• Which result is correct: $13\pi^2$ and $5\pi^2$? Apr 14, 2017 at 6:29
• They are the same just different way of writing it Apr 14, 2017 at 6:49
• Where does this integral come from?
– FDP
Apr 14, 2017 at 9:35
• Variation of Ahmed integral. Apr 14, 2017 at 9:40
• Where do the fraction $\dfrac{x+1}{x^2-x+2}$ come from? I mean if i replace it by, at random, $\dfrac{x+1}{x(x+2)}$ no such interresting result is found. Godess Namagiri gives you that integral in your dreams? :)
– FDP
Apr 14, 2017 at 10:08

One may observe that, for $x \in [0,1]$, $$\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)=\arctan\left({1\over \sqrt{x^2+2}}\right)+\arctan\left({x\over \sqrt{x^2+2}}\right)$$ then $$I:=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=I_1+I_2$$with \begin{align} I_1:&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left({1\over \sqrt{x^2+2}}\right)dx \\&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\left(\frac \pi2-\arctan\sqrt{x^2+2}\right)dx \\&=\frac \pi2\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}-\int_0^1 \frac{\arctan\sqrt{x^2+2}}{(1+x^2)\sqrt{x^2+2}}\:dx \\&=\frac{\pi^2}{12}-\frac{5\pi^2}{96} \\&=\frac{\pi^2}{32} \end{align} where the latter integral is Ahmed's evaluation.

On the other hand we have \begin{align} I_2:&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\cdot\arctan\left({x\over \sqrt{x^2+2}}\right)dx \\&=\int_1^\infty \frac{u}{(1+u^2)\sqrt{1+2u^2}}\cdot\arctan\left({1\over \sqrt{1+2u^2}}\right)du \quad (u=1/x) \\&=\frac12\int_1^\infty \frac{1}{(1+v)\sqrt{1+2v}}\cdot\arctan\left({1\over \sqrt{1+2v}}\right)dv \qquad (v=u^2) \\&=-\int_1^\infty \left[\arctan\left({1\over \sqrt{1+2v}}\right)\right]'\arctan\left({1\over \sqrt{1+2v}}\right)dv \\&=-\left[\frac12\cdot\arctan^2\left({1\over \sqrt{1+2v}}\right)\right]_1^\infty \\&=\frac{\pi^2}{72}. \end{align} Finally,

$$\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=\frac{\pi^2}{32}+\frac{\pi^2}{72}=\color{blue}{\frac{13\pi^2}{288}}$$

as announced.

• Nice answer. I understand now why Latte wasn't answering my question about the source of this integral, magicians don't reveal their tricks.
– FDP
Apr 14, 2017 at 15:17
• $\displaystyle \int \frac{\arctan\left(\frac{x}{\sqrt{x^2+2}}\right)}{(1+x^2)\sqrt{x^2+2}}dx=\dfrac{1}{2}\arctan^2\left(\frac{x}{\sqrt{x^2+2}}\right)+\text{constant}$
– FDP
Apr 14, 2017 at 15:41