Question:
$\frac{(x+1)(x+\log x)^{2}}{x}$
Solution:
$\frac{(x+1)(x+\log x)^{2}}{x}=\left(\frac{x+1}{x}\right)(x+\log x)^{2}=\left(1+\frac{1}{x}\right)(x+\log x)^{2}$
Let $(x+\log x)=t$
$\therefore\left(1+\frac{1}{x}\right) d x=d t$
$\Rightarrow \int\left(1+\frac{1}{x}\right)(x+\log x)^{2} d x=\int t^{2} d t$
$=\frac{t^{3}}{3}+\mathrm{C}$
$=\frac{1}{3}(x+\log x)^{3}+\mathrm{C}$
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