The value of ∫_0^1 8 log (1+x) 1+x^2 d x is

$\displaystyle \mathrm{I}=8 \int_0^1 \frac{\log (1+\mathrm{x})}{1+\mathrm{x}^2} \mathrm{dx}$I=8∫01​1+x2log(1+x)​dx $\displaystyle =8 \int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \sec ^2 \theta d \theta(\operatorname{let} x=\tan \theta)$=8∫04π​​1+tan2θlog(1+tanθ)​sec2θdθ(letx=tanθ) $\displaystyle =8 \int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta=8 \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta=8 \int_0^{\frac{\pi}{4}} \log 2 d \theta-8 \int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$=8∫04π​​log(1+tan(4π​−θ))dθ=8∫04π​​log(1+1+tanθ1−tanθ​)dθ=8∫04π​​log2dθ−8∫04π​​log(1+tanθ)dθ $=8 \log 2 \frac{\pi}{4}-I$=8log24π​−I $2 \mathrm{I}=2 \pi \log 2$2I=2πlog2 $\mathrm{I}=\pi \log 2$I=πlog2