Evaluate _1^ e 1 x(1+ x)^2 d x - Vidyadip

Question

Evaluate $\int_1^{ e } \frac{1}{x(1+\log x)^2} d x$

✓

Answer

Let $I =\int_1^{ e } \frac{1}{x(1+\log x)^2} d x$
Put $1+\log x=t$
$\therefore \frac{1}{x} d x= dt$
When $x=1, t=1+\log 1=1+0=1$
When $x=e, t=1+\log e=1+1=2$
$ \therefore I =\int_1^2 \frac{ dt }{ t ^2}$
$=\int_1^2 t ^{-2} dt$
$=\left[\frac{ t ^{-1}}{-1}\right]_1^2$
$=-\left[\frac{1}{ t }\right]_1^2 $
$=-\left(\frac{1}{2}-1\right)$
$\therefore I=-\left(\frac{-1}{2}\right)$
$=\frac{1}{2}$

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