TITLE: Evaluate $\int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$
QUESTION [12 upvotes]: Evaluate $\displaystyle \int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$

My effort:
\begin{align*}
I(a)&=\int^{2}_{0}\frac{\tan^{-1}(ax)}{1+4x}\mathrm dx\\
I'(a) &= \int^{2}_{0}\frac{x}{(1+4x)(1+a^2x^2)}\mathrm dx\\
I'(a) &= \frac{1}{4}\int^{2}_{0}\frac{(1+4x)-1}{(1+4x)(1+a^2x^2)}dx\\
I'(a) &= \frac{1}{4a}\tan^{-1}(2)-\frac{1}{4}\int^{2}_{0}\frac{1}{(1+4x)(1+a^2x^2)}dx
\end{align*}
Then how to proceed? Thank you.

REPLY [3 votes]: To calculate by series or to try to get a "closed form" ?  
I think the best is to use here the following formula:

$$\int\frac{\ln(x+z)}{x-a}\,dx = \text{Li}_2\left(\frac{x+z}{a+z}\right) - \ln(x+z)~\text{Li}_1\left(\frac{x+z}{a+z}\right) + C$$

It’s $~\displaystyle\tan x = \Im\ln(1+ix) = \frac{\pi}{2} + \Im\ln(x-i)~$ and we can set $~\displaystyle(a,z):=(-\frac{1}{4},-i)~$ . 
Then we get:
$\displaystyle \int\limits_0^2\frac{\arctan x}{1+4x}\,dx = \frac{\pi}{2}\int\limits_0^2\frac{dx}{1+4x} + \frac{1}{4}\Im\int\limits_0^2\frac{\ln(x-i)}{x-\left(-\frac{1}{4}\right)}\,dx$
$\displaystyle =\frac{\pi}{4}\ln 3 + \frac{1}{4}\Im\left(\text{Li}_2\left(\frac{8+i36}{17}\right) - \ln(2-i)~\text{Li}_1\left(\frac{8+i36}{17}\right) \right) $
$\displaystyle\hspace{1.8cm} - \frac{1}{4}\Im\left( \text{Li}_2\left(\frac{16+i4}{17}\right) - \ln(-i)~\text{Li}_1\left(\frac{16+i4}{17}\right) \right) $

Separation of real part and imaginary part:
$\displaystyle \ln(2-i) = \frac{\ln 5}{2} - i\arctan\frac{1}{2} \enspace , \enspace\enspace  \ln(-i) = -i\frac{\pi}{2}$
$\displaystyle \text{Li}_1\left(\frac{8+i36}{17}\right) = -\frac{1}{2}\ln\frac{81}{17} + i\arctan 4 \enspace , \enspace\enspace  \text{Li}_1\left(\frac{16+i4}{17}\right) = \frac{1}{2}\ln 17 + i\arctan 4$
Note: $\enspace$ For calculations with $~x>0~$ we can use $\displaystyle ~~\arctan x = \frac{\pi}{2} - \arctan\frac{1}{x}~$  .
$\displaystyle A:=\Im\text{Li}_2\left(\frac{16+i4}{17}\right) = \Im\text{Li}_2\left(\frac{\sqrt{272}}{17}e^{i\arctan\frac{1}{4}}\right)=\sum\limits_{k=1}^\infty\frac{\sin\left(k\arctan\frac{1}{4}\right)}{k^2}\left(\frac{\sqrt{272}}{17}\right)^k$
Using $~\displaystyle \text{Li}_2(z) = -\text{Li}_2\left(\frac{1}{z}\right) -\frac{\pi^2}{6}-\frac{1}{2}\ln^2(-z)~$ we get:
$\displaystyle \Im\text{Li}_2\left(\frac{8+i36}{17}\right) = \Im\text{Li}_2\left(\frac{\sqrt{1360}}{17}e^{i\arctan\frac{9}{2}}\right) = B - \left(\ln\frac{\sqrt{1360}}{17}\right)\left(-\pi + \arctan\frac{9}{2}\right)$
with $\enspace \displaystyle B := \sum\limits_{k=1}^\infty\frac{\sin\left(k\arctan\frac{9}{2}\right)}{k^2}\left(\frac{17}{\sqrt{1360}}\right)^k$
$\displaystyle R := \pi\ln\frac{36}{17} + \left(\ln\frac{85}{81}\right)\left(\arctan\frac{1}{4}\right) + \left(\ln\frac{80}{81}\right)\left(\arctan\frac{2}{9}\right)$ 
$A\approx 0.54772569304105608037669001217295320069571309$
$B\approx 0.46231444614312936610749685795341278946044980$
$R\approx 2.36624661042409606268346795278249163083184498$
$\displaystyle \int\limits_0^2\frac{\arctan x}{1+4x}\,dx = \frac{R}{8} + \frac{B-A}{4} \approx 0.2744280145785303292681352055429263510451648$