$\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int_{-1}^1 \log \left(\frac{2-x}{2+x}ight) \mathrm{d} x=0 \ & \because \log \left(\frac{2-x}{2+x}ight) 	ext { is an odd function. }\end{aligned}$

$\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=$

Solution