Question

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

Answer 1

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

Let

$\log x=t$

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

Also,

$\log x=t$

$\Rightarrow x=e^t$

Therefore,

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

$={\int }_1^2\frac{t}{e^t}dt$

$={\int }_1^{2}t{e}^{-t}dt$

$={[-t{e}^{-t}-\int (-e^{-t})dt]}_1^2$

$={[-t{e}^{-t}-e^{-t}]}_1^2$

$={\left[\frac{-t-1}{e^t}\right]}_1^2$

$=[\frac{-2-1}{e^2}-\frac{-1-1}{e^1}]$

$=[\frac{2}{e}-\frac{3}{e^2}]$

$=\frac{2e-3}{e^2}$

Therefore,

${\int }_1^2\frac{\log x}{x^2}dx=\frac{2e-3}{e^2}$

Hence the correct answer is $\frac{2e-3}{e^2}$.