Question
Evaluate the following integrals:
$\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$

Answer

Let $\text{I}=\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$
Here $\text{f(x)}=\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$\text{f}(-\text{x})=\log\Big(\frac{2+\text{x}}{2-\text{x}}\Big)$
$=-\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$=-\text{f(x)}$
Hence f(x) is an odd function,
Therefore,
$\text{I}=0$

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