Question:
Evaluate the following integrals:
$\int(x+1) \log x d x$
Solution:
Let $I=\int(x+1) \log x d x$
Using integration by parts,
$=\log x \int(x+1) d x-\int \frac{d}{d x} \log x \int(x+1) d x$
We know that, $\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}=\frac{1}{\mathrm{x}}$
$=\log x\left(\frac{x^{2}}{2}+x\right)-\int \frac{1}{x}\left(\frac{x^{2}}{2}+x\right) d x$
$=\left(\frac{x^{2}}{2}+x\right) \log x-\int \frac{x}{2} d x-\int d x$
$=\left(\frac{x^{2}}{2}+x\right) \log x-\frac{x^{2}}{4}-x+c$
$=\left(\frac{x^{2}}{2}+x\right) \log x-\left(\frac{x^{2}}{4}+x\right)+c$
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