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

Question

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

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Answer

$
\text { Let } \begin{aligned}
I & =\int_1^2 \frac{d x}{x(1+\log x)^2} \\
& =\int_1^2 \frac{1}{(1+\log x)^2} \cdot \frac{1}{x} d x
\end{aligned}
$
Put $1+\log x=t \quad \therefore \frac{1}{x} d x=d t$
When $x=1, t=1+\log 1=1+0=1$
When $x=2, t=1+\log 2$
$
\begin{aligned}
\therefore I & =\int_1^{1+\log 2} \frac{1}{t^2} d t=\int_1^{1+\log 2} t^{-2} d t \\
& =\left[\frac{t^{-1}}{-1}\right]_1^{1+\log 2}=-\left[\frac{1}{t}\right]_1^{1+\log 2} \\
& =-\left[\frac{1}{1+\log 2}-1\right] \\
& =-\left[\frac{1-(1+\log 2)}{1+\log 2}\right]=\frac{\log 2}{1+\log 2} .
\end{aligned}
$

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