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.
 A: Hint
Try$${x\over (1+4x)(1+a^2x^2)}={{-{4\over a^2+16}\over 1+4x}}+{{a^2x+4\over a^2+16}\over 1+a^2x^2}={1\over a^2+16}\left({{-{4}\over 1+4x}}+{{a^2x+4}\over 1+a^2x^2}\right)$$
A: When I tried to  simplify result, I found more elegant way to compute this integral.
We now that $\Im Log z= \phi=\arctan(\Im z/\Re z)$
Due to this I can rewrite the integral in more simple way
$$\int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx=\Im \int^{2}_{0}\frac{\log(1+ix)}{1+4x}\mathrm dx $$
It can be rewrite through logarith and dilogarithm by linear changing variable https://en.wikipedia.org/wiki/Spence%27s_function
$$I=\frac{\Im}{4} \left(-\text{Li}_2\left(\frac{16}{17}+\frac{4 i}{17}\right)+\text{Li}_2\left(\frac{8}{17}+\frac{36 i}{17}\right)+\log \left(\frac{9}{17}-\frac{36 i}{17}\right) \log (1+2 i)\right) $$
I am not a specialist in the polylogarithm, probably the imaginary parts of polilogarithm can be simplified.
The result for with additional parameter $$ \frac{\Im}{4} \left(\text{Li}_2\left(\frac{4 i-8 a}{a+4 i}\right)-\text{Li}_2\left(\frac{4 i}{a+4 i}\right)+\log \left(\frac{9 a}{a+4 i}\right) \log (1+2 i a)\right) $$
A: 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.
A: Yes,this integral seems hopeless from the point of view that we could get some simple closed form solution.
As a practical person who has to solve problems in real life where it is necessary to calculate not with great precision I use often the following approach (applied to given issue):
I'm trying to replace the integrand in the integral with a simpler expression so that the latter would differ as little as possible from the value of the original expression.
In this case I will replace $\arctan x$ with
$$\frac{\arctan 2}{26}x(23-5x)$$
The maximum deviation of this expression from $\arctan x$ in $[0,2]$ is less than $0.03$
Now, replacing $\arctan x$ in the integral with the expression I evaluate the integral and get
$$\frac{\arctan 2}{832}(308-97\ln 3)$$
The absolute error of this result is about $0.007$
This is even better result than expected. 
Now, noting that $\arctan 2$ and $\ln 3$ are both very close to $1$ I simplify the result to $\frac{211}{832}$
The absolute error of this last is still satisfactory from a practical point of view (about $0.021$)
In similar manner it is possible quickly estimate many difficult integrals (often easier than numerical integration procedures)
