# Evaluating $\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$ [duplicate]

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.

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## marked as duplicate by Marvis, 5PM, Douglas S. Stones, Henry T. Horton, amWhyJan 14 '13 at 2:39

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}
From the second expression in the first line: $\displaystyle{\int_{0}^{\pi/4}\ln\left(1 + \tan\left(t\right)\right)\,{\rm d}t = {1 \over 2}\left[\int_{0}^{\pi/4}\ln\left(1 + \tan\left(t\right)\right)\,{\rm d}t + \int_{0}^{\pi/4}\ln\left(1 + \tan\left({\pi \over 4} - t\right)\right)\,{\rm d}t \right]={1 \over 2}\left[\int_{0}^{\pi/4}\ln\left(1 + \tan\left(t\right)\right)\,{\rm d}t + \int_{0}^{\pi/4}\ln\left(2 \over 1 + \tan\left(t\right)\right)\,{\rm d}t \right]={1 \over 2}\int_{0}^{\pi/4}\ln\left(2\right)\,{\rm d} t={1 \over 8}\,\pi\ln\left(2\right)}$. – Felix Marin Mar 27 '14 at 7:47