Question

Integrate : ${\int }_1^2\frac{\ln x}{x^2}dx$

Answer 1

$\Rightarrow {\int }_1^2\frac{lnx}{x^2}dx$

let $\ln x=t$

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

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

$I={\int }_1^2\frac{lnx}{x^2}dx={\int }_{|n|=0}^{ln2}\frac{t{e}^t}{e^{2t}}dt={\int }_0^{ln2}{t}^{e^{-t}}dt$

Applying By parts gives

$I=(-t{e}^{-t}{)}_0^{ln2}+{\int }_0^{ln2}{e}^{-t}dt$

$I=-\ln 2\left(\frac{1}{2}\right)+(-e^{-t}{)}_0^{ln2}$

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

$\therefore {\int }_1^2\frac{\ln 2}{x^2}dx=\frac{1-\ln 2}{2}$