Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
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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