TITLE: Evaluation by methods of complex analysis $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm{dx}$
QUESTION [12 upvotes]: How would we evaluate the below integral by methods of complex analysis?
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm{dx}$$
I asked the question a while ago, but at that time I didn't specify this requirement.
Now, I'm only interested in a complex way. Thanks !!!
Sis.

REPLY [11 votes]: Well, there is a method which is not that complex. 
$$\int_0^1 \frac{\log(1+x)}{1+x^2}dx$$
$$\int_0^1 \left(\int_0^x \frac{1}{1+y}dy\right)\frac{1}{1+x^2}dx$$
$$\int_0^1 \int_0^x \frac{1}{1+y}dy\frac{1}{1+x^2}dx$$
Make a change of variables $y=ux$ in the inner integral:
$$\tag 1 \int_0^1 \int_0^1 \frac{x}{1+ux}\frac{1}{1+x^2}dudx$$
Partial fractions:
$${x \over {1 + xu}}{1 \over {1 + {x^2}}} = {1 \over {1 + {u^2}}}{x \over {1 + {x^2}}} + {u \over {1 + {u^2}}}{1 \over {1 + {x^2}}} - {u \over {1 + {u^2}}}{1 \over {1 + xu}}$$
Now, integrating the first two terms, which account to the same$^1$, gives that you integral is
$$\mathcal I=\frac \pi 4\log 2-\int_0^1\int_0^1 \frac u {1+u^2}\frac{1}{1+xu}dxdu$$
The latter integral is just our original integral, due to symmetry, as you see in $(1)$ 
This means that $$\mathcal I =\frac \pi 8 \log 2$$
as desired. You can see a simlar integral solved here with the very same method.
$1$: symmetry, once again.