Evaluación de $\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$

Question

Posible duplicado:
Evalúa la integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx$

$$\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$$

Me está costando mucho derivar la respuesta, $\frac{\pi}{8} \log(2)$ . He intentado la expansión de Taylor tanto del numerador como del denominador, ambas parecen demasiado complicadas e infructuosas.

Answer 1

Dejemos que $x = \tan(t)$ . Entonces obtenemos que $$\begin{align} I & = \int_0^1 \dfrac{\log(1+x)}{1+x^2} dx = \int_0^{\pi/4} \dfrac{\log(1+\tan(t))}{\sec^2(t)} \sec^2(t) dt\\ & = \int_0^{\pi/4} \log(\sin(t) + \cos(t)) dt - \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \int_0^{\pi/4} \log\left( \dfrac{\sin(t) + \cos(t)}{\sqrt{2}} \right) dt + \int_0^{\pi/4} \log(\sqrt{2}) dt- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \int_0^{\pi/4} \log(\cos(t-\pi/4)) dt + \int_0^{\pi/4} \log(\sqrt{2}) dt- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \underbrace{\int_{-\pi/4}^{0} \log(\cos(t)) dt}_{(t-\pi/4) \to t} + \int_0^{\pi/4} \dfrac{\log(2)}2 dt- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \underbrace{\int_{0}^{\pi/4} \log(\cos(t)) dt}_{t \to -t} + \dfrac{\pi}8 \log2- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \dfrac{\pi}8 \log 2 \end{align}$$