Question

Evaluate : ${\int }_1^2\frac{\log x}{x^2}dx$

Answer 1

${\int }_1^2\frac{\log x}{x^2}dx$

Let $v=\log x$

$dv=\frac{1}{x}dx$

$dv=x^{-2}dx$

$v=x^{-1}$

so:- ${\int }_1^2\frac{\log x}{x^2}dx$

$=-{\left[\frac{\log x}{x}\right]}_1^2+{\int }_1^2{x}^{-2}dx$

$=-\frac{1}{2}\log 2+\log 1+{[-\frac{1}{x}]}_1^2$

$=-\frac{1}{2}\log 2+0-\frac{1}{2}+1$

$=\frac{1}{2}-\frac{1}{2}\log 2$