Use ∫ ( ln ( x ) n = x ( ln x ) n − n ∫ ( ln x ) n − 1 d x {\displaystyle {\text{Use }}\int (\ln {(x)}^{n}=x(\ln {x})^{n}-n\int (\ln {x})^{n-1}dx}
∫ 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}}\\[2ex]&=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}}\\[2ex]&=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}}\\[2ex]&=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}}\\[2ex]&=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/69f953278018c3e639f388bee7d8ea8aace609f0)