Integral of $\frac{\ln(1+x^2)}{1+x^2}$ Compute $\int\frac{\ln(1+x^2)}{1+x^2}dx$
Attempt:
I tried to do Integration by Parts
$$u=\ln(1+x^2)\Rightarrow du=\frac{2xdx}{1+x^2}\\
dv=\frac{dx}{1+x^2}\Rightarrow v=\arctan x\\
\int\frac{\ln(1+x^2)}{1+x^2}dx=\ln(1+x^2)\arctan x-\int\frac{2x\arctan x dx}{1+x^2}$$
Since it seems that it not done in terms of elementary functions, what about the definite version:
$$\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx$$

$$\text{Let }I=\int\dfrac{\ln(1+x^2)}{1+x^2}\mathrm dx=\int\left(\underbrace{\dfrac{i\ln(1+x^2)}{2(x+i)}}_{I_1}-\underbrace{\dfrac{i\ln(1+x^2)}{2(x-i)}}_{I_2}\right)\mathrm dx$$

Solving $2/i\cdot I_1$:
$$\begin{bmatrix}u \\ \mathrm du\end{bmatrix}=\begin{bmatrix}x+i\\ \mathrm dx\end{bmatrix}$$ $$\begin{align}\int\dfrac{\ln(1+x^2)}{x+i}\mathrm dx&=\int\dfrac{\ln((u-i)^2+1)}{u}\mathrm du\\  & =\int\dfrac{\ln(u-2i)}{u}\mathrm du+\dfrac{1}{2}\ln^2(u)\\ & = \int\dfrac{\ln(iu/2+1)}{u}\mathrm du+\ln(-2i)\ln u+\dfrac{1}{2}\ln^2u \\ &= -\mathrm{Li}_2\left(-\dfrac{iu}{2}\right)+\dfrac{1}{2}\ln^2u+\ln(-2i)\ln(u)\\&=-\mathrm{Li}_{2}\left(-\dfrac{i(x+i)}{2}\right)+\dfrac{1}{2} \ln^2(x+i)+\ln(-2i)\ln(x+i)\tag1\end{align}$$

Similarly solve $2/i\cdot I_2$ to get the following result: $$\int\dfrac{\ln(1+x^2)}{x-i}\mathrm dx = -\mathrm{Li}_2\left(-\dfrac{i(x-i)}{2}\right)+\dfrac{1}{2}\ln^2(x-i)+\ln(2i)\ln(x-i)\tag2$$

Computing $i/2\cdot [(1)+(2)]\equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=\dfrac{i}{4}\left[\ln\mid x+i \mid \left(\ln\mid x+i\mid+2\ln(-2i)\right)-\ln\mid x-i\mid\left(\ln\mid x-i\mid +2\ln(2i)\right)\\  +2\mathrm{Li}_2\left(\dfrac{ix+1}{2}\right)-2\mathrm{Li}_2\left(-\dfrac{ix-1}{2}\right)\right]+C$$
