Evaluate the following integrals: ( )^2 dx

Question

Evaluate the following integrals:$\int(\log\text{x})^2\cdot\text{x dx}$

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Answer

Let $\text{I}=\int(\log\text{x})^2\text{x dx}$
Using integration by parts,
$=(\log\text{x})^2\int\text{x dx }-\int\Big(2(\log\text{x})\Big(\frac{1}{\text{x}}\Big)\int\text{x dx}\Big)\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-2\int(\log\text{x})\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{x}^2}{2}\Big)\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\int\text{x}(\log\text{x})\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\Big[\log\text{x}\int\text{x dx}-\int\Big(\frac{1}{\text{x}}\int\text{x dx}\Big)\text{dx}\Big]$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\Big[\frac{\text{x}^2}{2}\log\text{x}-\int\Big(\frac{1}{\text{x}}\times\frac{\text{x}^2}{2}\Big)\text{dx}\Big]$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\frac{\text{x}^2}{2}\log\text{x}+\frac{1}{2}\int\text{x dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\frac{\text{x}^2}{2}\log\text{x}+\frac{1}{4}\text{x}^2+\text{C}$
$\text{I}=\frac{\text{x}^2}{2}\Big[(\log\text{x})^2-\log\text{x}+\frac{1}{2}\Big]+\text{C}$

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