Question

${\int }_e^{e^2}[\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}]dx$

Answer 1

$I={\int }_e^{e^2}(\frac{1}{\log x}-\frac{1}{(\log x{)}^2})dx$

Put when

$\log x=t$ $x=e\Rightarrow t=1$

$\therefore x=e^t$ $x=e^2\Rightarrow t=2$

$dx=e^t.dt$

$={\int }_1^2(\frac{1}{t}-\frac{1}{t^2})e^t.dt$

Property $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+C$.

$f\left(t\right)=\frac{1}{t}\therefore f^{\prime }\left(t\right)=-\frac{1}{t^2}$

$={[e^t.\frac{1}{t}]}_1^2$

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