MCQ
$\int_1^{{e^2}} {\frac{{dx}}{{x{{(1 + \ln x)}^2}}}} =$
- ✓$2/3$
- B$1/3$
- C$3/2$
- D$\ln 2$
Let $(1 + \ln x) = t$
==> $dt = \frac{1}{x}dx$
Now, when $x = 1 \to {e^2}$, then $t = 1 \to 3$
$\therefore $ $I = \int_1^3 {\frac{{dt}}{{{t^2}}} = \left[ {\frac{{ - 1}}{t}} \right]_1^3= - \left[ {\frac{1}{3} - 1} \right]} = \frac{2}{3}$.
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