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The question was as follows: $$\int{\sqrt{\frac{e^x-1}{e^x+1}}}dx$$ Which I evaluated as follows: $$\int{\sqrt{\frac{e^x-1}{e^x+1}}}dx$$ $$=\int{\frac{e^x-1}{\sqrt{e^{2x}-1}}}dx$$ $$=\int{\frac{t-1}{t \sqrt{t^2-1}}}dt$$ (Taking $t=e^x$ and $dt= tdx$) $$=\int{\frac{t}{t\sqrt{t^2-1}}}dt-\int{\frac{1}{t\sqrt{t^2-1}}}dt$$ $$=\sin^{-1}(t)-\int{\frac{z}{z×t^2}}dz$$ (Taking $z^2=t^2-1$ and $dt=\cfrac{z}{t}dz$) $$=\sin^{-1}(t)-\int{\frac{1}{z^2+1}}dz$$ $$=\sin^{-1}(t)-\tan^{-1}(z)$$ $$=\sin^{-1}(e^x)-\tan^{-1}(\sqrt{e^{2x}-1})+C$$

But the answer given is: $$\ln (e^x+\sqrt{e^{2x}-1})-\sec^{-1}(e^x)+C$$

What did I do wrong and How to obtain the correct answer?

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    $\begingroup$ When you have split the integral into two parts, the first of these should be an inverse hyperbolic function, not inverse sine… $\endgroup$ Oct 5, 2023 at 7:37
  • $\begingroup$ Mathematica: $$2 \left(\tan ^{-1}\left(e^x-\sqrt{e^x-1} \sqrt{e^x+1}\right)+\tanh ^{-1}\left(\frac{1}{\sqrt{\frac{e^x-1}{e^x+1}}}\right)\right)$$ $\endgroup$ Oct 5, 2023 at 7:43

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The integrand can be rewritten as $\sqrt{\tanh \frac{x}{2}}$ so use the substitution $x = 2\tanh^{-1}u^2 $

$$I = \int \frac{4u^2}{1-u^4}du = \int\frac{2}{1-u^2}-\frac{2}{1+u^2}du = 2\tanh^{-1}u - 2\tan^{-1}u + C$$

Therefore the final answer is

$$\boxed{2\tanh^{-1}\sqrt{\tanh\frac{x}{2}}-2\tan^{-1}\sqrt{\tanh\frac{x}{2}}+C}$$

or equivalently

$$\boxed{2\tanh^{-1}\sqrt{\frac{e^x-1}{e^x+1}}-2\tan^{-1}\sqrt{\frac{e^x-1}{e^x+1}}+C}$$

This is equal to your given answer. For the arctan term, we have that

$$\alpha\in(0,1) \implies 2\tan^{-1}\alpha = \tan^{-1}\frac{2\alpha}{1-\alpha^2}$$

which means

$$2\tan^{-1}\sqrt{\frac{e^x-1}{e^x+1}} = \tan^{-1}\sqrt{e^{2x}-1} = \sec^{-1}e^x$$

by considering a right triangle with sides $\sqrt{e^{2x}-1}$ opposite, $1$ adjacent, and hypotenuse $e^x$. I will let you confirm the artanh term on your own.

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As I mentioned in the comments, you should have $$\operatorname {arcosh}(e^x)$$ instead of the inverse sine term. Apart from that, your answer is the same as the given answer since $$\operatorname{arcosh}(e^x)=\ln(e^x+\sqrt{e^{2x}-1})$$ and also $$\arctan(\sqrt{e^{2x}-1})=\sec^{-1}(e^x)$$

This is easily seen by using the identity $$1+\tan^2\theta=\sec^2\theta$$

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  • $\begingroup$ I believe I messed up the $\frac{1}{\sqrt{1-t^2}}$ and $\frac{1}{\sqrt{t^2-1}}$ $\endgroup$
    – Aurelius
    Oct 5, 2023 at 16:25
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    $\begingroup$ That’s basically all it was $\endgroup$ Oct 5, 2023 at 19:38

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