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Question Number 162513 by Lordose last updated on 30/Dec/21

$${\int }_0^{\mathrm{\infty }}\frac{\log \left(\mathrm{x}\right)}{(\mathrm{x}+1)(\mathrm{x}+9)}$$

Answered by abdullahoudou last updated on 30/Dec/21

$$x->\frac{9}{x}$$

Answered by Lordose last updated on 30/Dec/21

$$\mathrm{I}\stackrel{\mathrm{x}=\frac{9}{\mathrm{x}}}{=}9{\int }_0^{\mathrm{\infty }}\frac{\log \left(9\right)-\log \left(\mathrm{x}\right)}{{\mathrm{x}}^2(\frac{9}{\mathrm{x}}+1)(\frac{9}{\mathrm{x}}+9)}\mathrm{dx}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{2\log \left(3\right)-\log \left(\mathrm{x}\right)}{(\mathrm{x}+9)(\mathrm{x}+1)}\mathrm{dx}$$$$2\mathrm{I}=2\log \left(3\right){\int }_0^{\mathrm{\infty }}\frac{1}{(\mathrm{x}+9)(\mathrm{x}+1)}\mathrm{dx}$$$$2\mathrm{I}=\frac{\log \left(3\right)}{4}\log \left(\frac{\mathrm{x}+1}{\mathrm{x}+9}\right){\mid }_0^{\mathrm{\infty }}$$$$\mathrm{I}=\frac{{\log }^2\left(3\right)}{4}$$

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