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I have tried the substitution $x=\sinh u$. This got me to $\,\int \frac{\cosh^2u}{\sinh u}du$, and through an identity I got $\int(csch u+\sinh u)du$. But integrating this, and substituting $x$ back does not get me the correct answer. May someone tell me where I've gone wrong.

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  • $\begingroup$ Put $x=\tan y$. $\endgroup$ – Empty Oct 22 '15 at 2:53
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Hint

Even if $x=\sinh u$ does not look to be the best possible substitution, you can continue. $$I=\,\int \frac{\cosh^2(u)}{\sinh(u)}du=\,\int \frac{1+\sinh^2(u)}{\sinh(u)}du=\,\int \sinh(u)du+\,\int \frac{1}{\sinh(u)}du$$ Now, consider the last integrand$$\frac{1}{\sinh(u)}=\frac{\cosh^2(\frac u2)-\sinh^2(\frac u2)}{2\sinh(\frac u2)\cosh(\frac u2)}=\frac 12\Big(\frac{\cosh(\frac u2)}{\sinh(\frac u2)}-\frac{\sinh(\frac u2)}{\cosh(\frac u2)}\Big)$$ So, $$I=\,\int \sinh(u)du+\frac 12 \int \frac{\cosh(\frac u2)}{\sinh(\frac u2)}du-\frac 12 \int \frac{\sinh(\frac u2)}{\cosh(\frac u2)} du$$ For the last integrals make another change of variable $u=2v$ and arrive to $$I=\,\int \sinh(u)du+\int \frac{\cosh(v)}{\sinh(v)}dv- \int \frac{\sinh(v)}{\cosh(v)} dv$$

I am sure that you can take from here.

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Where things went wrong, if they did, cannot be determined without more information. What is written down so far is correct.

May I suggest a simpler way? Our function is $\frac{x\sqrt{1+x^2}}{x^2}$. Let $u^2=1+x^2$. Then $x\,dx=u\,du$, and we are integrating $\frac{u^2}{u^2-1}$.

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Hint:

Let $x = \tan{u} \implies dx = \sec^2{u} \ du$. Then,

$$\int \frac{\sqrt{ 1 + x^2}}{x}dx = \int\frac{du}{\sin{u}\cos^2u} = \int \csc {u} \sec^2{u} \ du = \int \csc \theta(1 + \tan^2\theta) \ du = \int \csc\theta \ du + \int \csc{u}\tan^2u \ du = \int \csc{u} \ du + \int \dfrac{\sin{u}}{\cos^2u} \ du.$$

It is essentially a simple rearrangement and substitution for different concepts to make this problem look a lot simpler than it seems.

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