Question
Integrate the function: $\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}$

Answer

Let $I = \int {\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}dx} $ ...(i)

Putting x + log x = t

$ \Rightarrow 1 + \frac{1}{x} = \frac{{dt}}{{dx}}$

$ \Rightarrow \frac{{x + 1}}{x} = \frac{{dt}}{{dx}}$

$\Rightarrow \left( {\frac{{x + 1}}{x}} \right)dx = dt$

$\therefore$ From eq. (i), $I = \int {{t^2}dx} $

$= \frac{{{t^3}}}{3} + c$

$= \frac{1}{3}{\left( {x + \log x} \right)^3} + c$

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