Explanation:
(D) : \(\int_{1}^{2} \frac{1}{x(1+\log x)^{2}} d x\)
Let, \(\quad 1+\log \mathrm{x}=\mathrm{t}\)
\(\frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\)
When, \(\mathrm{x}=1, \mathrm{t}=1\) and
When, \(x=2, t=1+\log 2\)
\(\therefore \quad \mathrm{I}=\int_{1}^{1+\log 2} \frac{1}{\mathrm{t}^{2}} \mathrm{dt}=\int_{1}^{1+\log 2} \mathrm{t}^{-2} \mathrm{dt}\)
\(=\left[\frac{\mathrm{t}^{-1}}{-1}\right]_{1}^{1+\log 2}=\left[-\frac{1}{\mathrm{t}}\right]_{1}^{1+\log 2}\)
\(=\left[\frac{-1}{1+\log 2}-\frac{-1}{1}\right]=\frac{-1}{1+\log 2}+1=\frac{-1+1+\log 2}{1+\log 2}\)
\(=\frac{\log 2}{1+\log 2}\)