How to show that: $\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)dx=\pi?$ I just wonder if this integral is correct.
How can we show that?

$$\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)\mathrm dx=\color{blue}\pi$$

Let $u=\tan x$
$$\int_{0}^{\infty}\left(\frac{1}{\ln u}+\frac{2-u^{4/3}}{1-u}\right)\frac{\mathrm du}{1+u^2}$$
Let $u=e^{y}$
$$\int_{1}^{\infty}\left(\frac{1}{y}+\frac{2-e^{4y/3}}{1-e^{y}}\right)\frac{e^y\mathrm du}{1+e^{2y}}$$
$$\frac{1}{2}\int_{-\infty}^{\infty}\left(\frac{1}{y}+\frac{2-e^{4y/3}}{1-e^{y}}\right)\frac{\mathrm dy}{\cosh y}$$
I can't continue...
 A: Elementary manipulations with the classical beta function reveal that 
$$I=\frac{\pi}{2}\int_0^1 \frac{2+\cos(\pi x/2)+3\sin(\pi x/2)}{1+\cos(\pi x/2)+\sin(\pi x/2)}\textrm{d} x =\pi.$$
A: Let $I$ be given by the integral
$$I=\int_0^{\pi/2}\left(\frac{1}{\log(\tan(x))}+\frac{2-\sqrt[3]{\tan^4(x)}}{1-\tan(x)}\right)\,dx\tag1$$
Enforce the substitution $x\mapsto \arctan(x)$ in $(1)$ to reveal
$$\begin{align}
I&=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{2-x^{4/3}}{x-1}\right)\,\frac{1}{1+x^2}\,dx\\\\
&=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{1}{x-1}\right)\,\frac{1}{1+x^2}\,dx+\int_0^\infty \left(\frac{1-x^{4/3}}{1-x}\right)\,\frac{1}{1+x^2}\,dx\tag2
\end{align}$$

The first integral on the right-hand side of $(2)$ is easy to evaluate by symmetry.  Let $J$ be given by
$$J=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{1}{x-1}\right)\,\frac{1}{1+x^2}\,dx\tag3$$
and enforce the substitution $x\mapsto 1/x$.  Then, we find that 
$$J=\int_0^\infty \left(-\frac{1}{\log(x)}-\frac{x}{1-x}\right)\,\frac{1}{1+x^2}\,dx\tag4$$
Adding $(3)$ and $(4)$, we find that 
$$2J=\int_0^\infty \frac{1}{1+x^2}\,dx=\frac\pi2$$
Hence, $J=\frac\pi4$.

We now proceed to evaluate the second integral on the right-hand side of $(2)$.  To that end, let $K$ be the integral given by
$$\begin{align}
K&=\int_0^\infty \left(\frac{1-x^{4/3}}{1-x}\right)\,\frac{1}{1+x^2}\,dx\\\\
&\overbrace{=}^{x\mapsto x^3}3\int_0^\infty \frac{1-x^4}{1-x^3}\frac{x^2}{1+x^6}\,dx\\\\
&=3\int_0^\infty \frac{x^2(1+x+x^2+x^3)}{(1+x+x^2)(1+x^6)}\,dx\\\\
&\overbrace{=}^{x\mapsto 1/x}3\int_0^\infty \frac{x(1+x+x^2+x^3)}{(1+x+x^2)(1+x^6)}\,dx\\\\
&=3\int_0^\infty \frac{x(x+1)}{(x^2+x+1)(x^4-x^2+1)}\tag5\\\\
&=\frac{3\pi}{4}
\end{align}$$
where $(5)$ can be evaluated using partial fraction expansion.

Putting it together yields the coveted equality
$$\int_0^{\pi/2}\left(\frac{1}{\log(\tan(x))}+\frac{2-\sqrt[3]{\tan^4(x)}}{1-\tan(x)}\right)\,dx=\pi$$
A: Under $x\to\frac{\pi}{2}-x$, one has
\begin{eqnarray}
I&=&\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)\mathrm dx\tag{1}\\
&=&\int_{0}^{\pi/2}\left(\frac{1}{\ln\cot x}+\frac{2-\sqrt[3]{\cot^4x}}{1-\cot x}\right)\mathrm dx\\
&=&\int_{0}^{\pi/2}\left(-\frac{1}{\ln\tan x}+\frac{\frac1{\sqrt[3]{\tan x}}-2\tan x}{1-\tan x}\right)\mathrm dx.\tag{2}
\end{eqnarray}
Adding (1) to (2), one has
\begin{eqnarray}
2I&=&\int_{0}^{\pi/2}\frac{\bigg(\frac1{\sqrt[3]{\tan x}}-\sqrt[3]{\tan^4x}\bigg)+2(1-\tan x)}{1-\tan x}\mathrm dx\\
&=&\pi+\int_{0}^{\pi/2}\frac{\frac1{\sqrt[3]{\tan x}}-\sqrt[3]{\tan^4x}}{1-\tan x}\mathrm dx
\end{eqnarray}
and the rest is the same as Mark Viola's answer under $u=\sqrt[3]{\tan x}$.
