Question
Integrate the function $(x^2 + 1) \log x$

Answer

$\int {\left( {{x^2} + 1} \right)\log xdx} $
$ = \int {\left( {\log x} \right)\left( {{x^2} + 1} \right)dx} $
[Applying product rule]
$= \log x\left( {\frac{{{x^3}}}{3} + x} \right) - \int {\frac{1}{x}\left( {\frac{{{x^3}}}{3} + x} \right)dx}$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \int {\left( {\frac{{{x^2}}}{3} + 1} \right)dx}$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {{x^2}dx - \int {1dx} } $
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\frac{{{x^3}}}{3} - x + c$
$= \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{{{x^3}}}{9} - x + c$

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