Question

Find:
${\int }_1^2\frac{\ell nx}{x^2}dx$

Answer 1

$u=\ln x\Rightarrow du=\frac{1}{x}dx$

$dv=\frac{1}{x^2}\Rightarrow v=\frac{-1}{x}$

${\int }_1^2\frac{\ln x}{x^2}dx={[\ln x\times \frac{-1}{x}]}_1^2-{\int }_1^2\frac{-1}{x}\times \frac{1}{x}dx$

$={\left[\frac{-\ln x}{x}\right]}_1^2-{\int }_1^2\frac{-1}{x^2}dx$

$=[\frac{-\ln 2}{2}-\frac{-\ln 1}{1}]+{\int }_1^2\frac{1}{x^2}dx$

$=\frac{-\ln 2}{2}+0+{\left[\frac{x^{-2+1}}{-2+1}\right]}_1^2$

$=\frac{-\ln 2}{2}+{\left[\frac{-1}{x}\right]}_1^2$

$=\frac{-\ln 2}{2}-[\frac{1}{2}-\frac{1}{1}]$

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

$=\frac{-\ln 2}{2}+\frac{1}{2}$

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