Prove $\int_0^1\frac{\arctan\sqrt{x^2+4}}{(x^2+2)\sqrt{x^2+4}}dx=(\frac14\arctan\sqrt2-\frac\pi8)\arctan\frac1{\sqrt2}+\frac\pi4\arctan\frac1{\sqrt5}$

Question

How to prove $$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)\;?$$ I came across this Ahmed integral on the site "Art of problem solving", and have found no proof so far. (These two problems seems to be related though). Any help will be appreciated!

Answer 1

To evaluate that integral we can use Feynman's trick: $$I=\int _0^1\frac{\arctan \left(\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$ $$I\left(a\right)=\int _0^1\frac{\arctan \left(a\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$ $$I'\left(a\right)=\int _0^1\frac{1}{\left(x^2+2\right)\left(a^2x^2+4a^2+1\right)}\:dx=\frac{1}{2a^2+1}\int _0^1\frac{1}{x^2+2}-\frac{a^2}{a^2x^2+4a^2+1}\:dx$$ $$=\frac{1}{2a^2+1}\left(\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}-\frac{a\arctan \left(\frac{a}{\sqrt{4a^2+1}}\right)}{\sqrt{4a^2+1}}\right)$$ Now lets integrate again: $$\int _1^{\infty }I'\left(a\right)\:da=\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}\int _1^{\infty }\frac{1}{2a^2+1}\:da-\underbrace{\int _1^{\infty }\frac{a\arctan \left(\frac{a}{\sqrt{4a^2+1}}\right)}{\sqrt{4a^2+1}\left(2a^2+1\right)}\:da}_{a=\frac{1}{x}}$$ $$\frac{\pi }{2}\int _0^1\frac{1}{\left(x^2+2\right)\sqrt{x^2+4}}dx-I\:=\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)}{2\sqrt{2}}\left(\frac{\pi \sqrt{2}}{2}-\sqrt{2}\arctan \left(\sqrt{2}\right)\right)-\int _0^1\frac{\arctan \left(\frac{1}{\sqrt{x^2+4}}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$ $$=\frac{\pi \arctan \left(\frac{1}{\sqrt{2}}\right)}{4}-\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{2}-\frac{\pi }{2}\int _0^1\frac{1}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx+\underbrace{\int _0^1\frac{\arctan \left(\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx}_{I}$$ $$-2I\:=\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{4}-\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{2}-\pi \underbrace{\int _0^1\frac{1}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx}_{t=\arctan \left(\frac{x}{\sqrt{x^2+4}}\right)}$$ $$I\:=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\int _0^{\arctan \left(\frac{1}{\sqrt{5}}\right)}\:dt$$ $$\boxed{I=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)}$$

This numerically agrees with Wolfram Alpha.

Answer 2

\begin{align} J&=\int_0^1 \frac{\arctan\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx\\ K&=\int_0^1 \int_0^1 \frac{1}{(x^2+2)(y^2+2)}dxdy \\ &=\int_0^1 \int_0^1 \frac{1}{4+x^2+y^2}\left(\frac{1}{2+x^2}+\frac{1}{2+y^2}\right)dxdy\\ &=2\int_0^1 \int_0^1 \frac{1}{(4+x^2+y^2)(2+x^2)}dxdy\\ &=2 \int_0^1 \left[\frac{\arctan\left(\frac{y}{\sqrt{4+x^2}}\right)}{(2+x^2)\sqrt{4+x^2}}\right]_{y=0}^{y=1} dx\\ &=2\int_0^1 \frac{\arctan\left(\frac{1}{\sqrt{4+x^2}}\right)}{(2+x^2)\sqrt{4+x^2}}dx\\ &=\pi \int_0^1 \frac{1}{(2+x^2)\sqrt{4+x^2}}dx-2J\\ &=\frac{\pi}{2} \left[\arctan\left(\frac{x}{\sqrt{4+x^2}}\right)\right]_0^1-2J\\ &=\frac{\pi}{2}\arctan\left(\frac{1}{\sqrt{5}}\right)-2J\\ \end{align}

On the other hand,

\begin{align}K&=\left(\int_0^1 \frac{1}{2+x^2}dx\right)^2\\ &=\left(\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_0^1\right)^2\\ &=\frac{1}{2}\arctan^2\left(\frac{1}{\sqrt{2}}\right) \end{align}

Therefore,

$\displaystyle \boxed{J=\frac{\pi}{4}\arctan\left(\frac{1}{\sqrt{5}}\right)-\frac{1}{4}\arctan^2\left(\frac{1}{\sqrt{2}}\right)}$

Answer 3

Let

$$ I(a)=\int_0^1 \frac{\arctan(a\sqrt{x^2+4})}{(x^2+2)\sqrt{x^2+4}}dx,I(0)=0,I=I(1)$$

\begin{align} I'(a)&=\int_0^1 \frac{1}{(x^2+4)[1+a^2(x^2+4)]}dx \\ &=\int_0^1 \frac{1}{x^2+4}dx-a^2\int_0^1 \frac{1}{1+a^2(x^2+4)}dx \\ &=\frac{1}{2} \arctan \frac{1}{2}-\frac{a}{\sqrt{1+4a^2}}\arctan \frac{a}{\sqrt{1+4a^2}} \\ \end{align}

Note

$$ \int \frac{a}{\sqrt{1+4a^2}}\arctan \frac{a}{\sqrt{1+4a^2}} da $$ $$ =\frac{1}{4} \int \arctan \frac{a}{\sqrt{1+4a^2}} d(\sqrt{1+4a^2}) $$

Using integration by parts,then I believe you can finish it

Answer 4

A generalized Ahmed integral takes the form \begin{align} \int_0^1 &\frac{\arctan\sqrt{x^2+a}}{\left(x^2+\frac a2\right) \sqrt{x^2+a}} \, dx =\int_0^1 \frac{\frac\pi2-\text{arccot}\sqrt{x^2+a}}{\left(x^2+\frac a2\right) \sqrt{x^2+a}} \, dx\\ & =\int_0^1 \int_0^1 \frac1{x^2+\frac a2}\bigg(\frac{\frac\pi2}{\sqrt{x^2+a}}-\frac{1}{x^2+y^2+a} \bigg)dy \ dx\\ &= \frac\pi a \text{arccot}\sqrt{1+a}- \frac1a\text{arccot}^2\sqrt{\frac a2} \end{align} which includes the original ($a=2$) as a special case $$\int_0^1\frac{\arctan\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}} \, dx=\frac{5\pi^2}{96}$$