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!
Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$
closed-formdefinite integralsintegration
Related Solutions
Complete answer now!$$I=\int_0^\infty \frac{{\arctan(x^2)}}{x^4+x^2+1}dx {\overset{x=\frac1{t}}=} \int_0^\infty \frac{\arctan\left(\frac{1}{t^2}\right)}{\frac{1}{t^4}+\frac{1}{t^2}+1}\frac{dt}{t^2}\overset{t=x}=\int_0^\infty \frac{{x^2\left(\frac{\pi}{2}-\arctan(x^2)\right)}}{x^4+x^2+1}dx $$ Now if we add the result with the original integral $I$ we get: $$2I=\frac{\pi}{2}\int_0^\infty \frac{x^2}{x^4+x^2+1}dx+\int_0^\infty \frac{(1-x^2)\arctan(x^2)}{x^4+x^2+1}dx$$ $$\Rightarrow I = \frac12 \cdot \frac{\pi}{2}\cdot \frac{\pi}{2\sqrt 3}-\frac12 \int_0^\infty \frac{(x^2-1)\arctan(x^2)}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt 3} -\frac12 J$$
Now in order to calculate $J\,$ we start by performing IBP: $$J=\int_0^\infty \frac{(x^2-1)\arctan\left(x^2\right)}{x^4+x^2+1}dx =\int_0^\infty \arctan(x^2) \left(\frac12 \ln\left(\frac{x^2-x+1}{x^2+x+1}\right)\right)'dx=$$ $$=\underbrace{\frac{1}{2}\ln\left(\frac{x^2-x+1}{x^2+x+1}\right)\arctan(x^2)\bigg|_0^\infty}_{=0}+\int_0^\infty \frac{x}{1+x^4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)dx$$ Substituting $x=\tan\left(t\right)$ and doing some simplifications yields: $$J=\int_0^\frac{\pi}{2} \frac{2\sin (2t)}{3+\cos(4t)}\ln\left(\frac{2+\sin (2t)}{2-\sin (2t)}\right)dt\overset{2t=x}=\int_0^\pi \frac{\sin x}{3+\cos(2x)}\ln\left(\frac{2+\sin x}{2-\sin x}\right)dx=$$ $$=2\int_0^\frac{\pi}{2}\frac{\sin x}{3+\cos(2x)}\ln\left(\frac{2+\sin x}{2-\sin x}\right)dx=\int_0^\frac{\pi}{2}\frac{\cos x}{1+\sin^2 x}\ln\left(\frac{2+\cos x}{2-\cos x}\right)dx =$$ $$=\frac12\int_0^\pi \frac{\cos x}{1+\sin^2 x}\ln\left(\frac{2+\cos x}{2-\cos x}\right)dx\overset{\large{\tan\left(\frac{x}{2}\right)=t}}=\int_0^\infty \frac{1-t^2}{t^4+6t^2+1}\ln\left(\frac{\color{blue}{t^2+3}}{\color{red}{3t^2+1}}\right)dt$$
Splitting the integral into two parts followed by the substitution $\,\displaystyle{t=\frac{1}{x}}\,$ in the second part gives: $$\int_0^\infty \frac{1-t^2}{t^4+6t^2+1}\ln(\color{red}{3t^2+1})dt =\int_0^\infty \frac{x^2-1}{x^4+6x^2+1}\ln\left(\color{red}{\frac{x^2+3}{x^2}}\right)dx$$ $$\Rightarrow J=\int_0^\infty \frac{1-x^2}{x^4+6x^2+1} \ln(\color{blue}{x^2+3})dx - \int_0^\infty \frac{1-x^2}{x^4+6x^2+1} {\left(\ln(\color{red}{x^2})-\ln(\color{red}{x^2+3})\right)}dx=$$ $$=2\int_0^\infty \frac{1-x^2}{x^4+6x^2+1}\ln\left(\color{purple}{\frac{x^2+3}{x}}\right)dx=2\int_0^\infty \left(\frac12\arctan\left(\frac{2x}{1+x^2}\right)\right)'\ln\left(\frac{x^2+3}{x}\right)dx=$$ $$=\underbrace{\arctan\left(\frac{2x}{1+x^2}\right)\ln\left(\frac{x^2+3}{x}\right)\bigg|_0^\infty}_{=0}-\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right)\left(\frac{2x}{x^2+3}-\frac{1}{x}\right)dx$$ $$\Rightarrow J=\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right)\frac{dx}{x}-\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right) \frac{2x}{x^2+3}dx=J_1-J_2$$
$$J_1=\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right)\frac{dx}{x}\overset{\large{x=\tan\left(\frac{t}{2}\right)}}=\int_0^\pi \frac{\arctan( \sin t)}{\sin t} dt\overset{t=x}=2\int_0^\frac{\pi}{2} \frac{\arctan( \sin x)}{\sin x} dx$$ In general, we have the following relation: $$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+(xy)^2} \Rightarrow \color{red}{\frac{\arctan(\sin x)}{\sin x}=\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}$$ $$J_1 = 2\color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=2\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=2\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$ $$=2\int_0^1 \frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}=\pi\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\boxed{\pi\ln\left(1+\sqrt 2\right)}$$
In order to evaluate $J_2$ we return the integral before was integrated by parts. $$J_2=2\int_0^\infty \arctan\left(\frac{2 x} {x^2 +1}\right)\frac{x}{x^2 +3}dx=2\int_0^{\infty}\frac{(x^2-1)\ln(x^2+3)}{x^4+6x^2+1} dx=$$ $$=(\sqrt 2+1)\int_0^{\infty} \frac{\ln(x^2+3)}{x^2+\left(\sqrt 2+1\right)^2} \ dx - (\sqrt 2-1)\int_0^{\infty} \frac{\ln(x^2+3)}{x^2+\left(\sqrt 2-1\right)^2} dx$$ Using the following identity that is valid for $a\ge 0, b>0$:$$\int_0^{\infty} \frac{\ln(x^2+a^2)}{x^2+b^2} \ dx = \frac{\pi}{b}\ln(a+b)$$ $$\Rightarrow J_2=\pi\ln\left(\frac{\sqrt{3}+\sqrt{2}+1}{\sqrt{3}+\sqrt{2}-1}\right)=\boxed{\frac{\pi} {2}\ln(2+\sqrt 3)}$$ So we found that:$$J=\boxed{\pi \ln(1+\sqrt 2) - \frac{\pi} {2} \ln(2+\sqrt 3)}\Rightarrow I= \large\boxed{\frac{\pi^2} {8 \sqrt 3}+\frac{\pi}{4}\ln(2+\sqrt 3)-\frac{\pi}{2} \ln(1+\sqrt 2)}$$
On the path of Zacky, the missing part...
Let,
\begin{align}I&=\int_0^{\frac{\pi}{2}}x^2\sqrt{\tan x}\,dx\\ J&=\int_0^{\frac{\pi}{2}}\frac{x^2}{\sqrt{\tan x}}\,dx\\ \end{align}
Perform the change of variable $y=\sqrt{\tan x}$,
\begin{align}I&=\int_0^{\infty}\frac{2x^2\arctan^2\left(x^2\right)}{1+x^4}\,dx\\\\ J&=\int_0^{\infty}\frac{2x^2\arctan^2\left(\frac{1}{x^2}\right)}{1+x^4}\,dx\\ \end{align}
\begin{align} \text{I+J}&=\int_0^{\infty}\frac{2x^2\left(\arctan\left(x^2\right)+\arctan\left(\frac{1}{x^2}\right)\right)^2}{1+x^4}\,dx-4\int_0^{\infty}\frac{x^2\arctan\left(x^2\right)\arctan\left(\frac{1}{x^2}\right)}{1+x^4}\,dx\\ &=\frac{\pi^2}{4}\int_0^{\infty}\frac{2x^2}{1+x^4}\,dx-4\int_0^{\infty}\frac{x^2\arctan\left(x^2\right)\arctan\left(\frac{1}{x^2}\right)}{1+x^4}\,dx\\ \end{align}
Perform the change of variable $y=\dfrac{1}{x}$,
\begin{align} \text{K}&=\int_0^{\infty}\frac{2x^2}{1+x^4}\,dx\\ &=\int_0^{\infty}\frac{2}{1+x^4}\,dx\\ \end{align}
Therefore,
\begin{align} \text{2K}=\int_0^{\infty}\frac{2\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+2}\,dx \end{align}
Perform the change of variable $y=x-\dfrac{1}{x}$,
\begin{align}\text{2K}&=2\int_{-\infty}^{+\infty}\frac{1}{2+x^2}\,dx\\ &=2\left[\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_{-\infty}^{+\infty}\\ &=2\times \frac{\pi}{\sqrt{2}} \end{align}
therefore,
\begin{align} \text{I+J}&=\frac{\pi^3}{4\sqrt{2}}-4\int_0^{\infty}\frac{x^2\arctan\left(x^2\right)\arctan\left(\frac{1}{x^2}\right)}{1+x^4}\,dx\\ \end{align}
Let $a>0$,
\begin{align} \text{K}_1(a)&=\int_0^{\infty}\frac{x^2}{a+x^4}\,dx\\ &=\frac{1}{a}\int_0^{\infty}\frac{x^2}{1+\left(a^{-\frac{1}{4}}x\right)^4}\,dx\\ \end{align}
Perform the change of variable $y=a^{-\frac{1}{4}}x$,
\begin{align} \text{K}_1(a)&=a^{-\frac{1}{4}}\int_0^{\infty}\frac{x^2}{1+x^4}\,dx\\ &=\frac{a^{-\frac{1}{4}}\pi}{2\sqrt{2}} \end{align}
In the same manner,
\begin{align} \text{K}_2(a)&=\int_0^{\infty}\frac{x^2}{1+ax^4}\,dx\\ &=\frac{a^{-\frac{3}{4}}\pi}{2\sqrt{2}} \end{align}
Since, for $a$ real,
\begin{align}\arctan a=\int_0^1 \frac{a}{1+a^2t^2}\,dt\end{align}
then,
\begin{align}\text{L}&=\int_0^{\infty}\frac{x^2\arctan\left(x^2\right)\arctan\left(\frac{1}{x^2}\right)}{1+x^4}\,dx\\ &=\int_0^{\infty}\left(\int_0^1 \int_0^1 \frac{x^2}{(1+u^2x^4)\left(1+\frac{v^2}{x^4}\right)(1+x^4)}\,du\,dv\right)\,dx\\ &=\\ &\int_0^{\infty}\left(\int_0^1\int_0^1 \left(\frac{x^2}{(1-u^2)(1-v^2)(1+x^4)}-\frac{x^2}{1-u^2v^2}\left(\frac{u^2}{(1-u^2)(1+u^2x^4)}+\frac{v^2}{(1-v^2)(v^2+x^4)}\right) \right)dudv\right)dx\\ &=\int_0^1\int_0^1 \left(\frac{\pi}{2\sqrt{2}(1-u^2)(1-v^2)}-\frac{1}{1-u^2v^2}\left(\frac{u^2\text{K}_2(u^2)}{1-u^2}+\frac{v^2\text{K}_1(v^2)}{1-v^2}\right)\right)dudv\\ &=\frac{\pi}{2\sqrt{2}}\int_0^1\int_0^1 \left(\frac{1}{(1-u^2)(1-v^2)}-\frac{1}{(1-u^2v^2)}\left(\frac{u^{\frac{1}{2}}}{1-u^2}+\frac{v^{\frac{3}{2}}}{1-v^2}\right)\right)dudv\\ &=\pi\int_0^1\left[\frac{\sqrt{v}\left(\text{ arctanh}\left(\sqrt{uv}\right)-\text{ arctan}\left(\sqrt{uv}\right)-\text{ arctanh}\left(uv\right)\right)+\arctan\left(\sqrt{u}\right)+\ln\left(\frac{\sqrt{1+u}}{1+\sqrt{u}}\right)}{2\sqrt{2}(1-v^2)}\right]_{u=0}^{u=1}\,dv\\ &=\frac{\pi}{2\sqrt{2}}\int_0^1\frac{\sqrt{v}\big(\text{ arctanh}\left(\sqrt{v}\right)-\text{ arctan}\left(\sqrt{v}\right)-\text{ arctanh}\left(v\right)\big)+\frac{\pi}{4}-\frac{1}{2}\ln 2}{1-v^2}\,dv\\ &=\frac{\pi}{2\sqrt{2}}\int_0^1\frac{\sqrt{v}\arctan\left(\frac{1-\sqrt{v}}{1+\sqrt{v}}\right)}{1-v^2}\,dv+\frac{\pi}{2\sqrt{2}}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right)\int_0^1 \frac{1-\sqrt{v}}{1-v^2}\,dv+\\ &\frac{\pi}{2\sqrt{2}}\int_0^1\frac{\sqrt{v}\ln\left(\frac{1+\sqrt{v}}{2}\right)}{1-v^2}\,dv-\frac{\pi}{4\sqrt{2}}\int_0^1\frac{\sqrt{v}\ln\left(\frac{1+v}{2}\right)}{1-v^2}\,dv \end{align}
Perform the change of variable $y=\dfrac{1-\sqrt{v}}{1+\sqrt{v}}$,
\begin{align}\text{R}_1&=\int_0^1\frac{\sqrt{v}\arctan\left(\frac{1-\sqrt{v}}{1+\sqrt{v}}\right)}{1-v^2}\,dv\\ &=\frac{1}{2}\int_0^1 \frac{(1-v)^2\arctan v}{v(1+v^2)}\,dv\\ &=\frac{1}{2}\int_0^1 \frac{\arctan v}{v}\,dv-\int_0^1 \frac{\arctan v}{1+v^2}\,dv\\ &=\frac{1}{2}\text{G}-\frac{1}{2}\Big[\arctan^2 v\Big]_0^1\\ &=\frac{1}{2}\text{G}-\frac{\pi^2}{32}\\ \text{R}_2&=\int_0^1 \frac{1-\sqrt{v}}{1-v^2}\,dv\\ &=\left[\ln\left(\frac{\sqrt{1+v}}{1+\sqrt{v}}\right)+\arctan\left(\sqrt{v}\right)\right]_0^1\\ &=\frac{\pi}{4}-\frac{1}{2}\ln 2\\ \end{align}
Perform the change of variable $y=\dfrac{1-\sqrt{v}}{1+\sqrt{v}}$,
\begin{align}\text{R}_3&=\int_0^1\frac{\sqrt{v}\ln\left(\frac{1+\sqrt{v}}{2}\right)}{1-v^2}\,dv\\ &=-\frac{1}{2}\int_0^1\frac{(1-v)^2\ln(1+v)}{v(1+v^2)}\,dv\\ &=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv-\frac{1}{2}\int_0^1 \frac{\ln(1+v )}{v}\,dv\\ &=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv-\frac{1}{4}\int_0^1 \frac{2v\ln(1-v^2)}{v^2}\,dv+\frac{1}{2}\int_0^1 \frac{\ln(1-v)}{v}\,dv\\ \end{align}
In the second integral perform the change of variable $y=v^2$,
\begin{align}\text{R}_3&=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv+\frac{1}{4}\int_0^1 \frac{\ln(1-v)}{v}\,dv\\ \end{align}
In the second integral perform the change of variable $y=1-v$,
\begin{align}\text{R}_3&=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv+\frac{1}{4}\int_0^1 \frac{\ln v}{1-v}\,dv\\ &=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv+\frac{1}{4}\times -\zeta(2)\\ &=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv-\frac{\pi^2}{24}\\ \end{align}
Perform the change of variable $y=\dfrac{1-v}{1+v}$,
\begin{align} \text{S}_1&=\int_0^1\frac{\ln(1+v)}{1+v^2}\,dv\\ &=\int_0^1\frac{\ln(\frac{2}{1+v})}{1+v^2}\,dv\\ &=\ln 2\int_0^1 \frac{1}{1+v^2}\,dv-\text{S}_1\\ &=\frac{\pi}{4}\ln 2-\text{S}_1 \end{align}
Therefore,
\begin{align} \text{S}_1&=\frac{\pi}{8}\ln 2\\ \text{R}_3&=\frac{\pi}{8}\ln 2-\frac{\pi^2}{24}\\ \end{align}
Perform the change of variable $y=\dfrac{1-\sqrt{v}}{1+\sqrt{v}}$,
\begin{align} \text{R}_4&=\int_0^1\frac{\sqrt{v}\ln\left(\frac{1+v}{2}\right)}{1-v^2}\,dv\\ &=\frac{1}{2}\int_0^1 \frac{(1-v)^2\ln\left(\frac{1+v^2}{(1+v)^2}\right)}{v(1+v^2)}\,dv\\ &=\frac{1}{2}\int_0^1 \frac{(1-v)^2\ln\left(1+v^2\right)}{v(1+v^2)}\,dv+2\text{R}_3\\ &=\frac{1}{2}\int_0^1\frac{\ln(1+v^2)}{v}\,dv-\int_0^1\frac{\ln(1+v^2)}{1+v^2}\,dv+\frac{\pi}{4}\ln 2-\frac{\pi^2}{12}\\ &=\frac{1}{2}\times \frac{1}{4}\zeta(2)-\int_0^1\frac{\ln(1+v^2)}{1+v^2}\,dv+\frac{\pi}{4}\ln 2-\frac{\pi^2}{12}\\ &=\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}-\int_0^1\frac{\ln(1+v^2)}{1+v^2}\,dv\\ &=\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}-\int_0^1\int_0^1\frac{v^2}{(1+v^2)(1+v^2t)}\,dt\,dv\\ &=\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}-\int_0^1 \left[\frac{\arctan\left(v\right)\sqrt{t}-\arctan\left(v\sqrt{t}\right)}{(t-1)\sqrt{t}}\right]_{v=0}^{v=1}\,dt\\ &=\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}-\int_0^1 \frac{\frac{\pi\sqrt{t}}{4}-\arctan\left(\sqrt{t}\right)}{(t-1)\sqrt{t}}\,dt\\ &=\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}+\int_0^1 \frac{\arctan\left(\frac{1-\sqrt{t}}{1+\sqrt{t}}\right)}{(1-t)\sqrt{t}}\,dt-\frac{\pi}{4}\int_0^1 \frac{\sqrt{t}-1}{(t-1)\sqrt{t}}\,dt\\ &=\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}+\int_0^1 \frac{\arctan\left(\frac{1-\sqrt{t}}{1+\sqrt{t}}\right)}{(1-t)\sqrt{t}}\,dt-\frac{\pi}{4}\Big[2\ln\left(1+\sqrt{t}\right)\Big]_0^1\\ &=\int_0^1 \frac{\arctan\left(\frac{1-\sqrt{t}}{1+\sqrt{t}}\right)}{(1-t)\sqrt{t}}\,dt-\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}\\ \end{align}
Perform the change of variable $y=\dfrac{1-\sqrt{t}}{1+\sqrt{t}}$,
\begin{align} \text{R}_4&=\int_0^1 \frac{\arctan t}{t}\,dt-\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}\\ &=\text{G}-\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}\\ \end{align}
Therefore,
\begin{align}L&=\frac{\pi}{2\sqrt{2}}\text{R}_1+\frac{\pi}{2\sqrt{2}}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right) \text{R}_2+\frac{\pi}{2\sqrt{2}}\text{R}_3-\frac{\pi}{4\sqrt{2}}\text{R}_4\\ &=\frac{\pi}{2\sqrt{2}}\left(\frac{\text{G}}{2}-\frac{\pi^2}{32}\right)+\frac{\pi}{2\sqrt{2}}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right)^2+\frac{\pi}{2\sqrt{2}}\left(\frac{\pi}{8}\ln 2-\frac{\pi^2}{24}\right)-\\ &\frac{\pi}{4\sqrt{2}}\left(\text{G}-\frac{\pi}{4}\ln 2-\frac{\pi^2}{16}\right)\\ &=\frac{\pi^3}{96\sqrt{2}}+\frac{\pi\ln^2 2}{8\sqrt{2}} \end{align}
Thus, \begin{align}\text{I+J}&=\frac{\pi^3}{4\sqrt{2}}-4\text{L}\\ &=\frac{\pi^3}{4\sqrt{2}}-4\left(\frac{\pi^3}{96\sqrt{2}}+\frac{\pi\ln^2 2}{8\sqrt{2}}\right)\\ &=\boxed{\frac{5\pi^3}{24\sqrt{2}}-\frac{\pi\ln^2 2}{2\sqrt{2}}} \end{align}
Best Answer
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.