<s>[INST] Evaluate $\int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$ 
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.
 [/INST] Firstly, do the integration by parts and and decompose a denominator
$$
\begin{align*}
& \int_0^2 \frac{\tan^{-1} (a x)}{1+4x} dx \\ = &\ \dfrac{\log 3 \tan^{-1} (2a)}{2}-\frac{a}{4}\int_0^2 \frac{\log(1+4x)}{1+a^2x^2} dx\\ = &\  \dfrac{\log 3 \tan^{-1} (2a)}{2}-\frac{a}{8}\int_0^2 \log(1+4x)(\frac{1}{1+iax}+\frac{1}{1-iax}) dx
\end{align*}$$
The obtained integrals can be rewrote through the dilogarithm $\text{Li}_2$ functions. The result have the following form
$$
\begin{align*}
I(a)&= \dfrac{\log 3 \tan^{-1} (2a)}{2}+\frac{i}{8}\left(\text{Li}_2\left(\frac{a}{a-4 i}\right)-\text{Li}_2\left(\frac{9 a}{a-4 i}\right)-\text{Li}_2\left(\frac{a}{a+4 i}\right) \\ + \text{Li}_2\left(\frac{9 a}{a+4 i}\right)-\log (9) \log \left(\frac{8 a+4 i}{-a+4 i}\right)+ \log (9) \log \left(\frac{-8 a+4 i}{a+4 i}\right)\right)
\end{align*}
$$
I think this answer can be simplified, but I do not want to do it.
</s>