$\int_{1}^{2} \frac{\mathrm{d} x}{x(1+\log x)^{2}}=$

Question

Aicharya · Accepted Answer

$I=\int_{1}^{2} \frac{d x}{x(1+\log x)}$ Put $\quad 1+\log x=t \Rightarrow \frac{1}{x} d x=d t$ When $x=1, t=1$ and when $x=2, t=1+\log 2$ $\begin{aligned} I &=\int_{1}^{1+\log 2} \frac{d t}{t}=[\log t]_{1}^{1+\log 2}=\log (1+\log 2)-\log 1 \ &=\log (1+\log 2) \end{aligned}$

$\int_{1}^{2} \frac{\mathrm{d} x}{x(1+\log x)^{2}}=$

Solution