Question

Evaluate: $\underset{0}{\overset{e^2}{\int }}\left[\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}\right]dx$

Answer 1

solution :-

$I={\int }_0^{{\theta }^2}[\frac{1}{logx}-\frac{1}{(logx{)}^2}]dx$

Now,take $logx=t$ Now at $x=0$ $t=\infty$

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

then $I={\int }^2(\frac{1}{t}-\frac{1}{t^2})etdt$

Now $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)$

$\therefore I=[e^t\frac{1}{t}{]}_{\infty }^2=\frac{e^2}{2}$