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Question Number 83898 by sahnaz last updated on 07/Mar/20

$$\int \frac{\mathrm{du}}{\sqrt{{\mathrm{u}}^2-1}-\mathrm{u}}$$

Commented by john santu last updated on 07/Mar/20

$$\mathrm{let}u=\sec \theta \Rightarrow \mathrm{du}=\sec \theta \tan \theta \mathrm{d}\theta$$

Commented by john santu last updated on 07/Mar/20

$$\int \frac{\sec \theta \tan \theta }{\tan \theta -\sec \theta }\mathrm{d}\theta =$$$$\int \frac{\sec \theta \tan \theta (\tan \theta +\sec \theta )}{-1}\mathrm{d}\theta$$$$-[\int \sec \theta \tan {}^2\theta +\tan \theta \sec {}^2\theta \mathrm{d}\theta ]$$$$=-\int \sec \theta (\sec {}^2\theta -1)\mathrm{d}\theta -\frac{1}{2}\tan {}^2\theta$$$$=-\int \sec \theta \mathrm{d}\left(\tan \theta \right)+\ln \mid \sec \theta +\tan \theta \mid -\frac{1}{2}\tan {}^2\theta$$$$$$

Commented by john santu last updated on 07/Mar/20

$$\mathrm{it}\mathrm{easy}\mathrm{to}\mathrm{solving}$$

Commented by mathmax by abdo last updated on 07/Mar/20

$$I=\int \frac{du}{\sqrt{u^2-1}-u}weusethechangementu=ch\left(t\right)\Rightarrow$$$$I=\int \frac{sh\left(t\right)dt}{sh\left(t\right)-ch\left(t\right)}=\int \frac{\frac{e^t-e^{-t}}{2}}{\frac{e^t-e^{-t}}{2}-\frac{e^t+e^{-t}}{2}}dt$$$$=\int \frac{e^t-e^{-t}}{e^t-e^{-t}-e^t-e^{-t}}dt=\int \frac{e^t-e^{-t}}{-2e^{-t}}dt=\int \frac{e^{2t}-1}{-2}dt$$$$=-\frac{1}{2}\int (e^{2t}-1)dt=-\frac{1}{4}{e}^{2t}+\frac{t}{2}+C$$$$t=argch\left(u\right)=ln(u+\sqrt{u^2-1})\Rightarrow e^{2t}={(u+\sqrt{u^2-1})}^2\Rightarrow$$$$I=-\frac{1}{4}{(u+\sqrt{u^2-1})}^2+\frac{1}{2}ln(u+\sqrt{u^2-1})+C$$

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