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I need to evaluate: $$\int \frac{\operatorname{csch}^2\sqrt u}{\sqrt{u}}$$

What I tried is write $$\operatorname{csch}^2\sqrt{u}=\coth^2\sqrt{u}-1$$ since the solution has a $\coth^2$ term, but I didn't go anywhere. I also tried to wrote out the formula for $\operatorname{csch}$ as $$\operatorname{csch} x=\dfrac{1}{\sinh x}=\dfrac{2}{e^x-e^{-x}}$$ still no clue about it... Any help? The solution is $-2\coth\sqrt{u}+c$

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    $\begingroup$ Substitute $x=\sqrt u$. $\endgroup$
    – user228113
    Commented Oct 9, 2016 at 23:01
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    $\begingroup$ Every integral of the form $\displaystyle\int\frac{f(\sqrt{u})}{\sqrt{u}}\, du$ can be turned into $2\displaystyle\int f(v)\, dv$ by setting $v = \sqrt{u}$. Then $\operatorname{csch}^{2}$ is a derivative you probably should know. $\endgroup$
    – Andrew D. Hwang
    Commented Oct 9, 2016 at 23:02

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Substitute $v=\sqrt{u}$. Then, $dv = \dfrac1 {2 \sqrt{u}} \, du$, and you have $2 \int \operatorname{csch}^2 v \,dv$.

Now, use the fact that $(\coth x)' = -\operatorname{csch}^2 x$.

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  • $\begingroup$ Should that be $(coth x)'=-csch^2 x$? $\endgroup$
    – Matata
    Commented Oct 9, 2016 at 23:13
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Hint: try $t=\sqrt{u}$ and $dt=\frac{du}{2\sqrt{u}}$

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