7.1 Integration By Parts/47

Prove $\int _{}^{}\left(\ln(x)^{n}\right)dx=x\left(\ln(x)^{n}\right)-n\int _{}^{}\left(\ln(x)^{n-1}\right)dx$

${\begin{aligned}u&=\ln(x)^{n}\quad dv=1dx\\[2ex]du&=1dx\qquad v=x\\[2ex]\end{aligned}}$

${\begin{aligned}\int _{}^{}\left(\ln(x)^{n}\right)dx&=x\ln(x)^{n}-\int _{}^{}\left((x{\frac {n\ln(x)^{n-1}}{x}})\right)dx\\[2ex]&=x\ln(x)^{n}-\int _{}^{}\left(n\ln(x)^{n-1}\right)dx\\[2ex]&=x\ln(x)^{n}-n\int _{}^{}\left(\ln(x)^{n-1}\right)dx\\[2ex]\end{aligned}}$