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Question Number 66728 by Tony Lin last updated on 19/Aug/19

$${\int }_1^{\mathrm{\infty }}\frac{1}{x\sqrt{x^2+1}}=?$$

Commented by mathmax by abdo last updated on 19/Aug/19

$$changementx=shtgive{\int }_1^{+\mathrm{\infty }}\frac{dx}{x\sqrt{1+x^2}}={\int }_{argsh\left(1\right)}^{+\mathrm{\infty }}\frac{ch\left(t\right)dt}{sh\left(t\right)ch\left(t\right)}$$$$=2{\int }_{ln(1+\sqrt{2})}^{+\mathrm{\infty }}\frac{dt}{e^t-e^{-t}}{=}_{e^t=u}2{\int }_{1+\sqrt{2}}^{+\mathrm{\infty }}\frac{1}{u-u^{-1}}\frac{du}{u}$$$$=2{\int }_{1+\sqrt{2}}^{+\mathrm{\infty }}\frac{du}{u^2-1}={\int }_{1+\sqrt{2}}^{+\mathrm{\infty }}\{\frac{1}{u-1}-\frac{1}{u+1}\}du={[ln\mid \frac{u-1}{u+1}\mid ]}_{1+\sqrt{2}}^{+\mathrm{\infty }}$$$$=-ln\mid \frac{1+\sqrt{2}-1}{1+\sqrt{2}+1}\mid =-ln\left(\frac{\sqrt{2}}{2+\sqrt{2}}\right)=ln\left(\frac{2+\sqrt{2}}{\sqrt{2}}\right)=ln(1+\sqrt{2})$$

Commented by Tony Lin last updated on 19/Aug/19

$$thankssir$$

Answered by Souvik Ghosh last updated on 19/Aug/19

$$\iff \mathrm{let}\mathrm{x}=\tan \mathrm{\Theta }\iff \mathrm{dx}=\sec {}^2\mathrm{\Theta }.\mathrm{d}\mathrm{\Theta }$$$$\mathrm{so}\mathrm{I}={\int }_{\pi /4}^{\pi /2}\frac{{\sec }^2\mathrm{\Theta }\mathrm{d}\mathrm{\Theta }}{\tan \mathrm{\Theta }\times \sec \mathrm{\Theta }}$$$$\mathrm{I}={\int }_{\pi /4}^{\pi /2}\mathrm{cosec}\mathrm{\Theta }\mathrm{d}\mathrm{\Theta }$$$$\mathrm{I}=-\ln [\sqrt{2}-1]$$$$$$

Answered by Souvik Ghosh last updated on 19/Aug/19

$$\iff \mathrm{let}\mathrm{x}=\tan \mathrm{\Theta }\iff \mathrm{dx}=\sec {}^2\mathrm{\Theta }.\mathrm{d}\mathrm{\Theta }$$$$\mathrm{so}\mathrm{I}={\int }_{\pi /4}^{\pi /2}\frac{{\sec }^2\mathrm{\Theta }\mathrm{d}\mathrm{\Theta }}{\tan \mathrm{\Theta }\times \sec \mathrm{\Theta }}$$$$\mathrm{I}={\int }_{\pi /4}^{\pi /2}\mathrm{cosec}\mathrm{\Theta }\mathrm{d}\mathrm{\Theta }$$$$\mathrm{I}=-\ln [\sqrt{2}-1]$$$$$$

Commented by Tony Lin last updated on 19/Aug/19

$$thankssir$$

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