Integration by parts 4

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Submitted by roamingfree on Fri, 02/05/2021 - 11:09
skills
You need to be able to apply the by parts formula
Question

Evaluate the integral

$$ \int x^2\ln(x^2) dx $$
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$\frac{1}{3}x^3\ln(x^2) - \frac{2}{9}x^3 + c$
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Let $u=\ln(x^2)$ then $u' = \frac{2}{x}$. Let $v' = x^2$, then $v = \frac{1}{3}x^3$. $$ \begin{array}{rcl} \int x^2\ln(x^2) dx &=& \frac{1}{3}x^3\ln(x^2) - \int \frac{1}{3}x^3\frac{2}{x}dx \\ &=& \frac{1}{3}x^3\ln(x^2) - \frac{2}{9}x^3 + c \end{array} $$