Use exercise 47 to evaluate ∫ ( ln x ) 3 d x {\displaystyle {\text{Use exercise 47 to evaluate}}\int (\ln {x})^{3}dx} Exercise 47: x ( ln x ) n − n ∫ ( ln x ) n − 1 d x {\displaystyle {\text{Exercise 47: }}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 ( x ) 3 − 3 [ ln 2 ( x ) ⋅ x − 2 ∫ ln ( x ) d x ] {\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-3\underbrace {\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}}\\&=x\ln(x)^{3}-3\left[\ln ^{2}{(x)}\cdot x-2\int \ln {(x)}dx\right]\\\end{aligned}}}
![{\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-3\underbrace {\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}}\\&=x\ln(x)^{3}-3\left[\ln ^{2}{(x)}\cdot x-2\int \ln {(x)}dx\right]\\\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6c302ba86e852b6cb60fa43ccfd32a224035057f)