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
$$\int_0^1  \frac{\ln(x+1)}{x^2+1} dx$$

Like Marvis, let us put $x=\tan\theta$
$$\int_0^1 \dfrac{\ln(1+x)}{1+x^2} dx  = \int_0^{\frac\pi4}  \ln(1+\tan\theta) d \theta$$
As $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
So, $$I=\int_0^{\frac\pi4}  \ln(1+\tan\theta) d\theta$$
$$=\int_0^{\frac\pi4}  \ln(1+\tan(\frac\pi4-\theta)) d\theta$$
$$=\int_0^{\frac\pi4}  \ln\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)d\theta$$
$$=\int_0^{\frac\pi4}  \ln\frac2{(1+\tan\theta)}d\theta$$
$$=\int_0^{\frac\pi4}  \{\ln2-\ln(1+\tan\theta)\}d\theta$$
$$=\int_0^{\frac\pi4}  \ln2- \int_0^{\frac\pi4}\ln(1+\tan\theta)d\theta$$
$$=\int_0^{\frac\pi4}  \ln2d\theta-I$$
So, $$2I=\int_0^{\frac\pi4}  \ln2d\theta$$
