TITLE: Evaluating $\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$
QUESTION [16 upvotes]: Possible Duplicate:
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx$ 

$$\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$$
I am having a hard time deriving the answer, $\frac{\pi}{8} \log(2) $.  I have tried Taylor expansion of both numerator and denominator, both seem too complicated and fruitless.

REPLY [19 votes]: Let $x = \tan(t)$. Then we get that
\begin{align}
I & = \int_0^1 \dfrac{\log(1+x)}{1+x^2} dx = \int_0^{\pi/4} \dfrac{\log(1+\tan(t))}{\sec^2(t)} \sec^2(t) dt\\
& = \int_0^{\pi/4} \log(\sin(t) + \cos(t)) dt - \int_0^{\pi/4} \log(\cos(t)) dt\\
& = \int_0^{\pi/4} \log\left( \dfrac{\sin(t) + \cos(t)}{\sqrt{2}} \right) dt + \int_0^{\pi/4} \log(\sqrt{2}) dt- \int_0^{\pi/4} \log(\cos(t)) dt\\
& = \int_0^{\pi/4} \log(\cos(t-\pi/4)) dt + \int_0^{\pi/4} \log(\sqrt{2}) dt- \int_0^{\pi/4} \log(\cos(t)) dt\\
& = \underbrace{\int_{-\pi/4}^{0} \log(\cos(t)) dt}_{(t-\pi/4) \to t} + \int_0^{\pi/4} \dfrac{\log(2)}2 dt- \int_0^{\pi/4} \log(\cos(t)) dt\\
& = \underbrace{\int_{0}^{\pi/4} \log(\cos(t)) dt}_{t \to -t} + \dfrac{\pi}8 \log2- \int_0^{\pi/4} \log(\cos(t)) dt\\
& = \dfrac{\pi}8 \log 2
\end{align}