MCQ
The value of $\int_1^{e^2} \frac{d x}{x(1+\log x)^2}$ is
  • $\frac{2}{3}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{3}{2}$
  • D
    $\frac{1}{2}$

Answer

Correct option: A.
$\frac{2}{3}$
(A)
Let $I =\int_1^{ e ^2} \frac{d x}{x(1+\log x)^2}$
Put $(1+\log x)= t \Rightarrow \frac{1}{x} d x= dt$
When $x=1, t =1$ and when $x= e ^2, t =3$
$\therefore \quad I=\int_1^3 \frac{ dt }{ t ^2}=\left[\frac{-1}{ t }\right]_1^3=-\left(\frac{1}{3}-1\right)=\frac{2}{3}$

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