$\int_1^{e^2} \frac{d x}{x(1+\log x)^2}=$

Question

Aicharya · Accepted Answer

$I=\int_1^{e^2} \frac{d x}{x(I+\log x)^2}$ Let $\quad 1+\log x=t$ $\frac{1}{x} d x=d t$ $\Rightarrow \quad I=\int_1^3 \frac{d t}{t^2}=\left[\frac{t^{-1}}{-1}\right]_1^3=\left(\frac{3^{-1}}{-1}+\frac{1^{-1}}{1}\right)$ $I=-\frac{1}{3}+1=\frac{2}{3}$

\(\int_1^{e^2} \frac{d x}{x(1+\log x)^2}=\)

Solution