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Question Number 31503 by abdo imad last updated on 09/Mar/18

$$find{\int }_2^{\sqrt{5}}\frac{dt}{t\sqrt{t^2-1}}.$$

Commented by abdo imad last updated on 16/Mar/18

$$ch.t=ch\left(x\right)giveI={\int }_{argch\left(2\right)}^{argch\left(\sqrt{5}\right)}\frac{shxdx}{ch\left(x\right)sh\left(x\right)}$$$$={\int }_{argch\left(2\right)}^{argch\left(\sqrt{5}\right)}\frac{dx}{\frac{e^x+e^{-x}}{2}}=2{\int }_{argch2)}^{argch\left(\sqrt{5}\right)}\frac{dx}{e^x+e^{-x}}$$$$letusethech.e^x=t\Rightarrow I=2{\int }_{e^{argch\left(2\right)}}^{e^{argch\left(\sqrt{5}\right)}}\frac{dt}{t(t+\frac{1}{t})}$$$$=2{\int }_{e^{argch\left(2\right)}}^{e^{argch\left(\sqrt{5}\right)}}\frac{dt}{t^2+1}=2{\left[arctan\left(t\right)\right]}_{e^{argch\left(2\right)}}^{e^{argch\left(\sqrt{5}\right)}}butweknow$$$$thatargch\left(x\right)=ln(x+\sqrt{x^2-1_{}})\Rightarrow$$$$argch\left(2\right)=ln(2+\sqrt{3})andargch\left(\sqrt{5}\right)=ln(\sqrt{5}+2)$$$$I=2{\left[arctan\left(t\right)\right]}_{2+\sqrt{3}}^{2+\sqrt{5}}=2(arctan(2+\sqrt{5})-arctan(2+\sqrt{3}))$$

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