Question

$\underset{1}{\overset{2}{\int }}\frac{1}{x{\left(1+\log x\right)}^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}1+\log x=t.\mathrm{Then},\frac{1}{x}dx=dt \\ \mathrm{When}x=1,t=1\mathrm{and}x=2,t=\left(1+\log 2\right) \\ \therefore I={\int }_1^2\frac{1}{x{\left(1+\log x\right)}^2}\mathit{d}x \\ \Rightarrow I={\int }_1^{\left(1+\log 2\right)}\frac{1}{t^2}dt \\ \Rightarrow I={\left[\frac{-1}{t}\right]}_1^{\left(1+\log 2\right)} \\ \Rightarrow I=-\frac{1}{\left(1+\log 2\right)}+1 \\ \Rightarrow I=\frac{\log 2}{\log 2+\log e} \\ \Rightarrow I=\frac{\log 2}{\log 2e} \end{gathered}$$