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

Let $x = \tan(t)$. We then have $dx = \sec^2(t)dt$ and $\dfrac1{1+x^2} = \dfrac1{1+ \tan^2(t)} = \dfrac1{\sec^2(t)}$.

Hence, $\dfrac{dx}{1+x^2} = dt$. \begin{align} \int_0^1 \dfrac{\ln(1+x)}{1+x^2} dx & = \int_0^{\pi/4} \ln(1+\tan(t)) dt = \int_0^{\pi/4} \ln(\cos(t)+\sin(t)) dt - \int_0^{\pi/4} \ln(\cos(t)) dt\\\ & = \int_0^{\pi/4} \ln(\sqrt{2}\cos(t-\pi/4)) dt - \int_0^{\pi/4} \ln(\cos(t)) dt\\\ & = \int_0^{\pi/4} \ln(\sqrt{2}) dt + \int_0^{\pi/4}\ln(\cos(t-\pi/4)) dt - \int_0^{\pi/4} \ln(\cos(t)) dt\\\ & = \dfrac{\pi \ln(2)}8 + \int_{-\pi/4}^0 \ln(\cos(t)) dt - \int_0^{\pi/4} \ln(\cos(t)) dt\\\ & = \dfrac{\pi \ln(2)}8 \,\,\,\,\,\,\,\,(\because \cos(t) \text{ is even}) \end{align}