Question

Evaluate the following definite integral:

${\int }_0^1\frac{1-x}{1+x}dx$

Answer 1

$I={\int }_0^1\frac{1-x}{1+x}dx={\int }_0^1\frac{2-(1+x)}{1+x}dx={\int }_0^1(\frac{2}{1+x}-1)dx={\left[2\log \right(x+1)-x]}_0^1$

$\Rightarrow I=(2\log 2-1)-(2\log 1-0)=2\log 2-1$