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:}}\qquad 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 {\displaystyle \int (\ln {x})^{3}dx=x(\ln {x})^{3}-3\int (\ln {x})^{2}dx} u = ( ln x ) 2 d v = d x {\displaystyle u=(\ln {x})^{2}\qquad dv=dx} d u = 2 1 x ln x d x v = x {\displaystyle du=2{\frac {1}{x}}\ln {x}dx\qquad v=x}
∫ ( ln x ) 3 d x = x ( ln x ) 3 − 3 ∫ ( ln x ) 2 d x = x ( ln x ) 3 − 3 [ x ( ln x ) 2 − 2 ∫ ( ln x ) d x ] = x ( ln x ) 3 − 3 [ x ( ln x ) 2 − 2 ( x ln x − ∫ 1 d x ) = x ( ln x ) 3 − 3 [ x ( ln x ) 2 − 2 x ln x + 2 x ] = x ( ln x ) 3 − 3 x ( ln x ) 2 + 6 x ln x − 6 x + c {\displaystyle {\begin{aligned}\int (\ln {x})^{3}dx&=x(\ln {x})^{3}-3\int (\ln {x})^{2}dx=x(\ln {x})^{3}-3[x(\ln {x})^{2}-2\int (\ln {x})dx]\\[2ex]&=x(\ln {x})^{3}-3[x(\ln {x})^{2}-2(x\ln {x}-\int 1dx)=x(\ln {x})^{3}-3[x(\ln {x})^{2}-2x\ln {x}+2x]\\[2ex]&=x(\ln {x})^{3}-3x(\ln {x})^{2}+6x\ln {x}-6x+c\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}\int (\ln {x})^{3}dx&=x(\ln {x})^{3}-3\int (\ln {x})^{2}dx=x(\ln {x})^{3}-3[x(\ln {x})^{2}-2\int (\ln {x})dx]\\[2ex]&=x(\ln {x})^{3}-3[x(\ln {x})^{2}-2(x\ln {x}-\int 1dx)=x(\ln {x})^{3}-3[x(\ln {x})^{2}-2x\ln {x}+2x]\\[2ex]&=x(\ln {x})^{3}-3x(\ln {x})^{2}+6x\ln {x}-6x+c\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/33a3c229d886df59bee5b83696b8d235af5a7596)