Show that $\int_{0}^{1} \frac{\ln|x+1|}{x^2+1}dx= \frac\pi8 \ln2$ using a suitable substitution

Let $x=\tan u$ then $\mathrm dx=\sec ^2 u\,\mathrm du$ . $$ \int^\frac{π}{4}_0 \frac{\ln(|1+ \tan u|)}{1+ \tan ^2 u} \sec^2 u\,\mathrm du = \int^\frac{π}{4}_0 \ln(|1+\tan u|)\,\mathrm du$$ Substituting $\frac{π}{4} - u$ in place of $u$, \begin{gather*} \int^\frac{π}{4}_0 \ln\left|1+\tan\left(\frac{π}{4}-u\right)\right|\,\mathrm du = \int^\frac{π}{4}_0 \ln\left|\frac{2}{1+\tan u}\right|\,\mathrm du\\ =\int^\frac{π}{4}_0 \ln2\,\mathrm du - \int^\frac{π}{4}_0 \ln|1+\tan u|\,\mathrm du=\int^\frac{π}{4}_0 \ln|1+\tan u|\,\mathrm du \end{gather*} On looking at the last two equalities we get the result.