Question

Find the value of ${\int }_0^1\log xdx$

Answer 1

${\int }_0^1\log xdx={\int }_0^1(x+\log x+\frac{x+\log x}{x}-\frac{x+logx}{x}-x)dx$

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

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

$={[x\log x-x]}_0^1$

$=1\log 1-1-(0-0)$

$=0-1$

${\int }_0^1\log xdx=-1$