How to compute $\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$ I have been asked to compute the integral $$\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$$
I have been told that it converges, and we only need its value.
I tried the substitutions $ u=\tan(x)$, $\;\; v^2=\tan(x) $ .
I thought about by parts integration, but it becomes more complicated.
I thank you in advance for an idea .
 A: $$\begin{align}
\int_{0}^{\frac{\pi}{2}}\frac{\arctan\left(\sqrt{\tan\left(x\right)}\right)}{\tan\left(x\right)}dx &= \frac{1}{2}\int_{0}^{\pi}\frac{\arctan\left(\sqrt{\tan\left(\frac{x}{2}\right)}\right)}{\tan\left(\frac{x}{2}\right)}dx
\\&\stackrel{(1)}{=}\int_{0}^{\infty}\frac{\arctan\left(\sqrt{t}\right)}{t\left(1+t^{2}\right)}dt
\\&=2\int_{0}^{\infty}\frac{\arctan u}{u\left(1+u^{4}\right)}du
\\&\stackrel{(2)}{=}2\int_{0}^{\infty}\int_{0}^{1}\frac{1}{1+u^{2}t^{2}}\frac{1}{1+u^{4}}dtdu
\\&\stackrel{(3)}{=}2\int_{0}^{1}\int_{0}^{\infty}\frac{1}{1+u^{2}t^{2}}\frac{1}{1+u^{4}}dudt
\\&\stackrel{(4)}{=}2\int_{0}^{1}\frac{1}{1+t^{4}}\int_{0}^{\infty}\left(\frac{1}{1+u^{4}}+\frac{t^{4}}{t^{2}u^{2}+1}-\frac{t^{2}u^{2}}{1+u^{4}}\right)dudt
\\&=2\int_{0}^{1}\frac{1}{1+t^{4}}\left(\frac{\pi}{2\sqrt{2}}+\frac{\pi t^{3}}{2}-\frac{\pi t^{2}}{2\sqrt{2}}\right)dt
\\&= \frac{\pi}{\sqrt{2}}\int_{0}^{1}\left(\frac{1-t^{2}}{1+t^{4}}\right)dt+\pi\int_{0}^{1}\frac{t^{3}}{1+t^{4}}dt
\\&=\frac{\pi}{\sqrt{2}}\frac{\operatorname{arccoth}\left(\sqrt{2}\right)}{\sqrt{2}}+\frac{\pi\log\left(2\right)}{4}
\\&=\frac{\pi}{4}\log\left(6+4\sqrt{2}\right) = \frac{\pi}{2}\log\left(2+\sqrt{2}\right)
\end{align}$$
Where:
$(1)$ is the Weirstrass Subsitution
$(2)$ is using the fact $\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $
$(3)$ is interchanging integrals via Fubini's Theorem (our integrand is positive and decays rapidly)
$(4)$ is a partial fraction decomposition
A: Here is an alternative solution to the proposed integral

We first introduce the following parameter:$$I(k)=\int_{0}^{\pi/2}\frac{\arctan(k\sqrt{\tan  t})}{\tan t}dt$$ Clearly the desired integral is $I(1)$. We now differentiate under the integral sign: $$I'(k) = \int_{0}^{\pi /2}\frac{dt}{\sqrt{\tan t}\left(k^2\tan t+1\right)}$$
Letting $\tan(t) \mapsto t$ we have
$$I'(k)=\int_{0}^{\infty}\frac{dt}{\sqrt{t}\left(k^2t+1\right)\left(1+t^2\right)}$$ Via an easy partial fraction decomposition followed by some basic integration, we conclude $$I'(k)=\frac{2\pi k^3-\sqrt{2}\pi\left(k^2-1\right)}{2\left(k^4+1\right)}$$ Note the resemblance to the integrand on the third-from-last line in my other answer—this is not accidental. The same (not very difficult) integrations show
$$I(k)=\frac{\pi}{2}\log\left(k^2+k\sqrt{2}+1\right)$$ So we find $$I(1) = \frac{\pi}{2}\log\left(2+\sqrt{2}\right)$$
