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Question Number 58319 by salahahmed last updated on 21/Apr/19

$$\int x^n{\left(\ln x\right)}^{n}dx$$

Answered by Smail last updated on 22/Apr/19

$$Byparts$$$$u={\left(lnx\right)}^n\Rightarrow u^{\prime }=n\frac{{\left(lnx\right)}^{n-1}}{x}$$$$v^{\prime }=x^n\Rightarrow v=\frac{x^{n+1}}{n+1}$$$$I=\int x^n{\left(lnx\right)}^{n}dx=\frac{x^{n+1}}{n+1}{\left(lnx\right)}^n-\frac{n}{n+1}\int x^n{\left(lnx\right)}^{n-1}dx+c_1$$$$Byparts$$$$u={\left(lnx\right)}^{n-1}\Rightarrow u^{\prime }=(n-1)\frac{{\left(lnx\right)}^{n-2}}{x}$$$$v^{\prime }=x^n\Rightarrow v=\frac{x^{n+1}}{n+1}$$$$I=\frac{x^{n+1}}{n+1}{\left(lnx\right)}^n-\frac{n}{n+1}(\frac{x^{n+1}}{n+1}{\left(lnx\right)}^{n-1}-\frac{n-1}{n+1}\int x^n{\left(lnx\right)}^{n-2}dx)+c_2$$$$=x^{n+1}(\frac{{\left(lnx\right)}^n}{n+1}-\frac{n{\left(lnx\right)}^{n-1}}{{(n+1)}^2})+\frac{n(n-1)}{{(n+1)}^2}\int x^n{\left(lnx\right)}^{n-2}dx+c_2$$$$=\frac{x^{n+1}}{n+1}({\left(lnx\right)}^n-\frac{n{\left(lnx\right)}^{n-1}}{n+1}+\frac{n(n-1){\left(lnx\right)}^{n-2}}{{(n+1)}^2}-...+{(-1)}^{n-1}\frac{n!\left(lnx\right)}{{(n+1)}^{n-1}})+\frac{{(-1)}^{n}n!}{{(n+1)}^n}\int x^{n}dx+c_n$$$$=\frac{x^{n+1}}{n+1}(\frac{n!{\left(lnx\right)}^n}{(n-0)!}-\frac{n!{\left(lnx\right)}^{n-1}}{(n-1)!(n+1)}+...+{(-1)}^{n-1}\frac{n!\left(lnx\right)}{(n-(n-1))!{(n+1)}^{n-1}})+\frac{{(-1)}^{n}n!}{0!{(n+1)}^n}\left(\frac{x^{n+1}}{n+1}\right)+C$$$$\int x^n{\left(lnx\right)}^{n}dx=\frac{n!x^{n+1}}{n+1}\underset{i=0}{\sum ^n}\frac{{(-1)}^i{\left(lnx\right)}^{n-i}}{(n-i)!{(n+1)}^i}+C$$

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