How to evaluate $\int_0^1 \frac{\ln(x+1)}{x^2+1} dx$ This problem appears at the end of Trig substitution section of Calculus by Larson. I tried using trig substitution but it was a bootless attempt
$$\int_0^1  \frac{\ln(x+1)}{x^2+1} dx$$

Substitute $x=\tan(\theta)$ the integral then reduces to 
$$ \int_0^{\frac{\pi}{4}} \ln(1+\tan(\theta)) d\theta \\
 = \int_0^{\frac{\pi}{4}} \ln(\frac{\sqrt{2}\cos(\pi/4-\theta)}{\cos(\theta)}) d\theta \\
 = \int_0^{\frac{\pi}{4}} \ln(\sqrt{2})d\theta \\
= \frac{\pi \ln(2)}{8}
$$
