Failed to parse (syntax error): {\displaystyle \text{Use } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx \\ \text{ to evaluate } \int(\ln{x})^3dx}
∫ ln ( x ) 3 d x = x ln ( x ) 3 − 3 ∫ ln ( x ) 2 d x ⏟ u = ln 2 ( x ) d v = d x d u = 2 ln ( x ) x d x v = x = x ln 3 ( x ) − 3 [ ln 2 ( x ) ⋅ x − 2 ∫ ln ( x ) d x ] = x ln 3 ( x ) − 3 x ln 2 ( x ) + 6 ∫ ln ( x ) d x ⏟ u = ln ( x ) d v = d x d u = 1 x d x v = x = x ln 3 ( x ) − 3 x ln 2 ( x ) + 6 [ ln ( x ) ⋅ x − ∫ d x ] = x ln 3 ( x ) − 3 x ln 2 ( x ) + 6 x ln ( x ) − 6 x + C {\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-\underbrace {3\int \ln(x)^{2}dx} _{\begin{aligned}u&=\ln ^{2}{(x)}&dv&=dx\\[0.6ex]du&={\tfrac {2\ln {(x)}}{x}}dx&v&=x\end{aligned}}\\[1ex]&=x\ln ^{3}(x)-3\left[\ln ^{2}{(x)}\cdot x-2\int \ln {(x)}dx\right]\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+\underbrace {6\int \ln {(x)}dx} _{\begin{aligned}u&=\ln {(x)}&dv&=dx\\[0.6ex]du&={\tfrac {1}{x}}dx&v&=x\end{aligned}}\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+6\left[\ln {(x)}\cdot x-\int dx\right]\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+6x\ln {(x)}-6x+C\end{aligned}}}
![{\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-\underbrace {3\int \ln(x)^{2}dx} _{\begin{aligned}u&=\ln ^{2}{(x)}&dv&=dx\\[0.6ex]du&={\tfrac {2\ln {(x)}}{x}}dx&v&=x\end{aligned}}\\[1ex]&=x\ln ^{3}(x)-3\left[\ln ^{2}{(x)}\cdot x-2\int \ln {(x)}dx\right]\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+\underbrace {6\int \ln {(x)}dx} _{\begin{aligned}u&=\ln {(x)}&dv&=dx\\[0.6ex]du&={\tfrac {1}{x}}dx&v&=x\end{aligned}}\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+6\left[\ln {(x)}\cdot x-\int dx\right]\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+6x\ln {(x)}-6x+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/1fe5ef5657d897738f08183930482187f1a3aff3)