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Question Number 58319 by salahahmed last updated on 21/Apr/19
∫x^n (lnx)^n dx
$$\int{x}^{{n}} \left(\mathrm{ln}{x}\right)^{{n}} {dx} \\ $$
Answered by Smail last updated on 22/Apr/19
By parts u=(lnx)^n ⇒u′=n(((lnx)^(n−1) )/x) v′=x^n ⇒v=(x^(n+1) /(n+1)) I=∫x^n (lnx)^n dx=(x^(n+1) /(n+1))(lnx)^n −(n/(n+1))∫x^n (lnx)^(n−1) dx+c_1 By parts u=(lnx)^(n−1) ⇒u′=(n−1)(((lnx)^(n−2) )/x) v′=x^n ⇒v=(x^(n+1) /(n+1)) I=(x^(n+1) /(n+1))(lnx)^n −(n/(n+1))((x^(n+1) /(n+1))(lnx)^(n−1) −((n−1)/(n+1))∫x^n (lnx)^(n−2) dx)+c_2 =x^(n+1) ((((lnx)^n )/(n+1))−((n(lnx)^(n−1) )/((n+1)^2 )))+((n(n−1))/((n+1)^2 ))∫x^n (lnx)^(n−2) dx+c_2 =(x^(n+1) /(n+1))((lnx)^n −((n(lnx)^(n−1) )/(n+1))+((n(n−1)(lnx)^(n−2) )/((n+1)^2 ))−...+(−1)^(n−1) ((n!(lnx))/((n+1)^(n−1) )))+(((−1)^n n!)/((n+1)^n ))∫x^n dx+c_n =(x^(n+1) /(n+1))(((n!(lnx)^n )/((n−0)!))−((n!(lnx)^(n−1) )/((n−1)!(n+1)))+...+(−1)^(n−1) ((n!(lnx))/((n−(n−1))!(n+1)^(n−1) )))+(((−1)^n n!)/(0!(n+1)^n ))((x^(n+1) /(n+1)))+C ∫x^n (lnx)^n dx=((n!x^(n+1) )/(n+1))Σ_(i=0) ^n (((−1)^i (lnx)^(n−i) )/((n−i)!(n+1)^i ))+C
$${By}\:{parts} \\ $$$${u}=\left({lnx}\right)^{{n}} \Rightarrow{u}'={n}\frac{\left({lnx}\right)^{{n}−\mathrm{1}} }{{x}} \\ $$$${v}'={x}^{{n}} \Rightarrow{v}=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$${I}=\int{x}^{{n}} \left({lnx}\right)^{{n}} {dx}=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\left({lnx}\right)^{{n}} −\frac{{n}}{{n}+\mathrm{1}}\int{x}^{{n}} \left({lnx}\right)^{{n}−\mathrm{1}} {dx}+{c}_{\mathrm{1}} \\ $$$${By}\:{parts} \\ $$$${u}=\left({lnx}\right)^{{n}−\mathrm{1}} \Rightarrow{u}'=\left({n}−\mathrm{1}\right)\frac{\left({lnx}\right)^{{n}−\mathrm{2}} }{{x}} \\ $$$${v}'={x}^{{n}} \Rightarrow{v}=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$${I}=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\left({lnx}\right)^{{n}} −\frac{{n}}{{n}+\mathrm{1}}\left(\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\left({lnx}\right)^{{n}−\mathrm{1}} −\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}}\int{x}^{{n}} \left({lnx}\right)^{{n}−\mathrm{2}} {dx}\right)+{c}_{\mathrm{2}} \\ $$$$={x}^{{n}+\mathrm{1}} \left(\frac{\left({lnx}\right)^{{n}} }{{n}+\mathrm{1}}−\frac{{n}\left({lnx}\right)^{{n}−\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right)+\frac{{n}\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int{x}^{{n}} \left({lnx}\right)^{{n}−\mathrm{2}} {dx}+{c}_{\mathrm{2}} \\ $$$$=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\left(\left({lnx}\right)^{{n}} −\frac{{n}\left({lnx}\right)^{{n}−\mathrm{1}} }{{n}+\mathrm{1}}+\frac{{n}\left({n}−\mathrm{1}\right)\left({lnx}\right)^{{n}−\mathrm{2}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−…+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{n}!\left({lnx}\right)}{\left({n}+\mathrm{1}\right)^{{n}−\mathrm{1}} }\right)+\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}+\mathrm{1}\right)^{{n}} }\int{x}^{{n}} {dx}+{c}_{{n}} \\ $$$$=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\left(\frac{{n}!\left({lnx}\right)^{{n}} }{\left({n}−\mathrm{0}\right)!}−\frac{{n}!\left({lnx}\right)^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)!\left({n}+\mathrm{1}\right)}+…+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{n}!\left({lnx}\right)}{\left({n}−\left({n}−\mathrm{1}\right)\right)!\left({n}+\mathrm{1}\right)^{{n}−\mathrm{1}} }\right)+\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{0}!\left({n}+\mathrm{1}\right)^{{n}} }\left(\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)+{C} \\ $$$$\int{x}^{{n}} \left({lnx}\right)^{{n}} {dx}=\frac{{n}!{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}} \left({lnx}\right)^{{n}−{i}} }{\left({n}−{i}\right)!\left({n}+\mathrm{1}\right)^{{i}} }+{C} \\ $$

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