Question

Find the value of : ${\int }_1^2\frac{dx}{x(1+\log x{)}^2},$

• A. $2/3$
• B. $1/3$
• C. $3/2$
• D. $\frac{\log 2}{1+\log 2}$

Answer 1

The correct option is D $\frac{\log 2}{1+\log 2}$

${\int }_1^2\frac{dx}{x{(1+logx)}^2}$

Put $1+logx=t\Rightarrow \frac{dx}{x}=dt$

lower limit $=1+log1=1$

upper limit $=1+log2$

$\therefore$ $I={\int }_1^{1+log2}\frac{dt}{t^2}$

$={-t}^{-1}={\left[\frac{-1}{t}\right]}_1^{1+log2}=\frac{-1}{1+log2}-\left(\frac{-1}{1}\right)$

$=1-\frac{1}{1+log2}$

$=\frac{log2}{1+log2}$ [D]