Question

What is the integral of limits 1 to 2 of function 1/x(1+logx)2

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{I}={\int }_1^2\frac{dx}{x{\left(1+\log x\right)}^2} \\ \mathrm{put}1+\log x=t \\ \Rightarrow \frac{1}{x}dx=dt \\ \mathrm{as},x\to 1,t\to 1 \\ \mathrm{as}x\to 2,t\to 1+\log 2 \\ \mathrm{Now}\mathrm{I}={\int }_1^{1+\log 2}\frac{dt}{t^2}=-{\left[\frac{1}{t}\right]}_1^{1+\log 2}=-\left[\frac{1}{1+\log 2}-1\right] \\ =1-\frac{1}{1+\log 2} \\ =\frac{1+\log 2-1}{1+\log 2} \\ =\frac{\log 2}{1+\log 2} \end{gathered}$$