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 = 1 2 ( x + y ) x = 1 2 ( u + v ) v = 1 2 ( x − y ) y = 1 2 ( u − v ) {\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-3\underbrace {\int \ln(x)^{2}dx} _{\begin{aligned}u&={\tfrac {1}{\sqrt {2}}}(x+y)&x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)&y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-3\underbrace {\int \ln(x)^{2}dx} _{\begin{aligned}u&={\tfrac {1}{\sqrt {2}}}(x+y)&x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)&y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8e4efbd472fe4b94a28fc8302fca01bec4520f70)