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

Question

Aicharya · Accepted Answer

$\int_{\frac{-1}{2}}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) \mathrm{d} x=0\left[\because \int_{-a}^a f(x) \mathrm{d} x=0\right.$ if $f(x)$ is odd $]$

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

Solution