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x-n-lnx-dx-




Question Number 25837 by mubeen897@hotmail.com last updated on 15/Dec/17
∫(x^n lnx)dx
$$\int\left({x}^{{n}} {lnx}\right){dx} \\ $$
Answered by kaivan.ahmadi last updated on 15/Dec/17
u=lnx⇒du=(dx/x)  dv=x^n dx⇒v=(x^(n+1) /(n+1))  now we have  ∫x^n lnxdx=uv−∫vdu=  ((x^(n+1)  )/(n+1))lnx−∫(x^n /(n+1))dx=(x^(n+1) /(n+1))lnx−(x^(n+1) /((n+1)^2 ))+C  =(x^(n+1) /(n+1))(lnx−(1/(n+1)))+C
$$\mathrm{u}=\mathrm{lnx}\Rightarrow\mathrm{du}=\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\mathrm{dv}=\mathrm{x}^{\mathrm{n}} \mathrm{dx}\Rightarrow\mathrm{v}=\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\mathrm{x}^{\mathrm{n}} \mathrm{lnxdx}=\mathrm{uv}−\int\mathrm{vdu}= \\ $$$$\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} \:}{\mathrm{n}+\mathrm{1}}\mathrm{lnx}−\int\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\mathrm{lnx}−\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{C} \\ $$$$=\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\left(\mathrm{lnx}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)+\mathrm{C} \\ $$
Commented by mubeen897@hotmail.com last updated on 16/Dec/17
thanks(−;
$${thanks}\left(−;\right. \\ $$

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