Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 125276 by Study last updated on 09/Dec/20

$$\underset{0}{\stackrel{1}{\int }}\frac{x^9-1}{lnx}dx=???$$

Answered by Dwaipayan Shikari last updated on 09/Dec/20

$$I\left(a\right)={\int }_0^1\frac{x^a-1}{logx}dx$$$$\Rightarrow I^{\prime }\left(a\right)={\int }_0^1{x}^{a}dx=\frac{1}{a+1}$$$$I\left(a\right)=\int \frac{1}{a+1}=log(a+1)+C$$$$a=0C=0$$$$I\left(a\right)=log(a+1)\Rightarrow I\left(9\right)=log\left(10\right)$$

Answered by mathmax by abdo last updated on 09/Dec/20

$$\mathrm{I}={\int }_0^1\frac{{\mathrm{x}}^9-1}{\mathrm{lnx}}\mathrm{dx}\mathrm{changement}\mathrm{lnx}=-\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{e}}^{-\mathrm{t}}$$$$\mathrm{I}=-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-9\mathrm{t}}-1}{-\mathrm{t}}(-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}=-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-10\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{\mathrm{t}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}-{\mathrm{e}}^{-10\mathrm{t}}}{\mathrm{t}}\mathrm{dt}\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}-{\mathrm{e}}^{-10\mathrm{t}}}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}\mathrm{with}[\mathrm{x}>0$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-{\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{-\mathrm{t}}-{\mathrm{e}}^{-10\mathrm{t}}){\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{-(\mathrm{x}+10)\mathrm{t}}-{\mathrm{e}}^{-(\mathrm{x}+1)\mathrm{t}})\mathrm{dt}$$$$={[-\frac{1}{\mathrm{x}+10}{\mathrm{e}}^{-(\mathrm{x}+10)\mathrm{t}}+\frac{1}{\mathrm{x}+1}{\mathrm{e}}^{-(\mathrm{x}+1)\mathrm{t}}]}_0^{\mathrm{\infty }}$$$$=\frac{1}{\mathrm{x}+10}-\frac{1}{\mathrm{x}+1}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\ln \left(\frac{\mathrm{x}+10}{\mathrm{x}+1}\right)+\mathrm{C}\mathrm{due}\mathrm{to}\mathrm{continuity}$$$$\mathrm{\exists }\mathrm{m}>0/\mid \mathrm{f}\left(\mathrm{x}\right)\mid ⩽\mathrm{m}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}=\frac{\mathrm{m}}{\mathrm{x}}\to 0(\mathrm{x}\to +\mathrm{\infty })\Rightarrow \mathrm{c}=0\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\ln \left(\frac{\mathrm{x}+10}{\mathrm{x}+1}\right)\mathrm{and}\mathrm{I}=\mathrm{f}\left(0\right)=\ln \left(10\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com