## University Calculus: Early Transcendentals (3rd Edition)

$$\int16x^3(\ln x)^2dx=4x^4(\ln x)^2-2x^4\ln x+\frac{x^4}{2}+C$$
$$A=\int16x^3(\ln x)^2dx$$ Use Reduction Formula 117, which states that $$\int x^n(\ln ax)^mdx=\frac{x^{n+1}(\ln ax)^m}{n+1}-\frac{m}{n+1}\int x^n(\ln ax)^{m-1}dx$$ for $n=3$, $m=2$ and $a=1$: $$A=16\Big[\frac{x^4(\ln x)^2}{4}-\frac{2}{4}\int x^3\ln xdx\Big]$$ $$A=4x^4(\ln x)^2-8\int x^3\ln xdx$$ Apply Formula 117 one more time, this time for $n=3$, $m=1$ and $a=1$: $$A=4x^4(\ln x)^2-8\Big[\frac{x^4\ln x}{4}-\frac{1}{4}\int x^3dx\Big]$$ $$A=4x^4(\ln x)^2-2x^4\ln x+2\int x^3dx$$ $$A=4x^4(\ln x)^2-2x^4\ln x+2\frac{x^4}{4}+C$$ $$A=4x^4(\ln x)^2-2x^4\ln x+\frac{x^4}{2}+C$$