Question
Integrate the function: $\frac{{{{\left( {1 + \log x} \right)}^2}}}{x}$
Putting $1 + \log x = t$
$ \Rightarrow \frac{1}{x} = \frac{{dt}}{{dx}}$
$\Rightarrow \frac{{dx}}{x} = dt$
$\therefore$ From eq. (i), $I = \int {{t^2}dt} $
$= \frac{{{t^3}}}{3} + c$
$= \frac{1}{3}{\left( {1 + \log x} \right)^3} + c$
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