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Question Number 36182 by prof Abdo imad last updated on 30/May/18

$$calculate{\int }_1^{+\mathrm{\infty }}\frac{dt}{t\sqrt{1+t^2}}$$

Commented by maxmathsup by imad last updated on 15/Aug/18

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

Answered by sma3l2996 last updated on 31/May/18

$$L={\int }_1^{+\mathrm{\infty }}\frac{dt}{t\sqrt{1+t^2}}$$$$lett=sinhx\Rightarrow dt=coshxdx$$$$L={\int }_{ln(\sqrt{2}+1)}^{+\mathrm{\infty }}\frac{coshx}{sinhx\sqrt{1+{sinh}^{2}x}}dx$$$$={\int }_{ln(\sqrt{2}+1)}^{+\mathrm{\infty }}\frac{dx}{sinh\left(x\right)}=2{\int }_{ln(\sqrt{2}+1)}^{+\mathrm{\infty }}\frac{e^x}{e^{2x}-1}dx$$$$u=e^x\Rightarrow du=e^{x}dx$$$$L=2{\int }_{\sqrt{2}+1}^{+\mathrm{\infty }}\frac{du}{u^2-1}=2{\int }_{\sqrt{2}+1}^{+\mathrm{\infty }}\frac{du}{(u+1)(u-1)}$$$$={\int }_{\sqrt{2}+1}^{+\mathrm{\infty }}(\frac{1}{u-1}-\frac{1}{u+1})du$$$$={[ln\mid \frac{u-1}{u+1}\mid ]}_{\sqrt{2}+1}^{+\mathrm{\infty }}=ln\mid \frac{\sqrt{2}+2}{\sqrt{2}}\mid =$$$$L=ln(\sqrt{2}+1)$$$$$$

Answered by ajfour last updated on 31/May/18

$$lett=\tan \theta ;then$$$$I={\int }_{\pi /4}^{\pi /2}\frac{\sec {}^2\theta d\theta }{\left(\tan \theta \right)\sec \theta }$$$$I={\int }_{\pi /4}^{\pi /2}\mathrm{cosec}\theta d\theta$$$$=\ln \mid \frac{\mathrm{cosec}\left(\pi /2\right)-\cot \left(\pi /2\right)}{\mathrm{cosec}\left(\pi /4\right)-\cot \left(\pi /4\right)}\mid$$$$=\ln \left(\frac{1}{\sqrt{2}-1}\right)=\ln (1+\sqrt{2}).$$$$$$

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