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Question Number 116557 by Bird last updated on 04/Oct/20

$$calculate{\int }_0^{\mathrm{\infty }}\frac{ln(1+x(1+t^2)}{1+t^2}dt$$ $$withx>0$$ $$2)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{ln(2+t^2)}{1+t^2}dt$$ $$and{\int }_0^{\mathrm{\infty }}\frac{ln(3+2t^2)}{1+t^2}dt$$

Answered by mindispower last updated on 05/Oct/20

$$f\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{ln(1+x(1+t^2))}{1+t^2}dt$$ $$f^{\prime }\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{1+x(1+t^2)}$$ $$={\int }_0^{\mathrm{\infty }}\frac{dt}{(1+x)(1+{\left(\frac{t}{\sqrt{1+x}}\right)}^2)}$$ $$=\frac{1}{\sqrt{1+x}}{\left[{\tan }^{-1}\left(\frac{t}{\sqrt{1+x}}\right)\right]}_0^{\mathrm{\infty }}$$ $$=\frac{\pi }{2\sqrt{1+x}}$$ $$f\left(0\right)=0$$ $$f\left(x\right)={\int }_0^x\frac{\pi }{2\sqrt{1+x}}=\pi \sqrt{1+x}$$ $${\int }_0^{\mathrm{\infty }}\frac{ln(2+t^2)}{1+t^2}dt=f\left(1\right)=\pi \sqrt{2}$$ $${\int }_0^{\mathrm{\infty }}\frac{ln(3+2t^2)}{1+t^2}dt=f\left(2\right)=\pi \sqrt{3}$$ $$$$

Commented bymathmax by abdo last updated on 05/Oct/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented bymathmax by abdo last updated on 05/Oct/20

$$(1+\mathrm{x})(1+{\left(\frac{\mathrm{t}}{\sqrt{1+\mathrm{x}}}\right)}^2)=(1+\mathrm{x})\{1+\frac{{\mathrm{t}}^2}{1+\mathrm{x}}\}=1+\mathrm{x}+{\mathrm{t}}^2\ne 1+\mathrm{x}(1+{\mathrm{t}}^2)!$$ $$$$

Answered by mathmax by abdo last updated on 05/Oct/20

$$1)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{\ln (1+\mathrm{x}(1+{\mathrm{t}}^2))}{1+{\mathrm{t}}^2}\mathrm{dt}\mathrm{we}\mathrm{have}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+\mathrm{x}(1+{\mathrm{t}}^2)}$$ $$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{xt}}^2+\mathrm{x}+1}=\frac{1}{\mathrm{x}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{\mathrm{x}+1}{\mathrm{x}}}{=}_{\mathrm{t}=\sqrt{\frac{\mathrm{x}+1}{\mathrm{x}}}\mathrm{u}}\frac{1}{\mathrm{x}}.\frac{\mathrm{x}}{\mathrm{x}+1}{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{u}}^2+1}\frac{\sqrt{\mathrm{x}+1}}{\sqrt{\mathrm{x}}}\mathrm{du}$$ $$=\frac{1}{\sqrt{\mathrm{x}}\sqrt{\mathrm{x}+1}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}}\sqrt{\mathrm{x}+1}}+\mathrm{c}$$ $${=}_{\sqrt{\mathrm{x}}=\mathrm{z}}\int \frac{2\mathrm{zdz}}{\mathrm{z}\sqrt{{\mathrm{z}}^2+1}}+\mathrm{c}=2\int \frac{\mathrm{dz}}{\sqrt{{\mathrm{z}}^2+1}}+\mathrm{c}$$ $$=2\ln (\mathrm{z}+\sqrt{1+{\mathrm{z}}^2})+\mathrm{c}=2\ln (\sqrt{\mathrm{x}}+\sqrt{1+\mathrm{x}})+\mathrm{c}$$ $$\mathrm{f}\left(0\right)=0=2\ln \left(1\right)+\mathrm{c}=\mathrm{c}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=2\ln (\sqrt{\mathrm{x}}+\sqrt{1+\mathrm{x}})$$ $$2){\int }_0^{\mathrm{\infty }}\frac{\ln (2+{\mathrm{t}}^2)}{1+{\mathrm{t}}^2}\mathrm{dt}=\mathrm{f}\left(1\right)=2\ln (1+\sqrt{2})$$ $${\int }_0^{\mathrm{\infty }}\frac{\ln (3+{2\mathrm{t}}^2)}{1+{\mathrm{t}}^2}\mathrm{dt}=\mathrm{f}\left(2\right)=2\ln (\sqrt{2}+\sqrt{3})$$

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