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How to evaluate:

$$\int \dfrac{\operatorname d \! x}{\sinh^4 x}$$

I tried to split it in $\int \frac{1}{\sinh^2x}\frac{1}{\sinh^2x}$ and then integrate by parts, but it's seems to complicate the whole thing.

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Hint:

$$\begin{align} \int\operatorname{csch}^4{x}\,\mathrm{d}x &=\int\operatorname{csch}^2{x}\left(-1+\coth^2{x}\right)\,\mathrm{d}x\\ &=-\int\operatorname{csch}^2{x}\,\mathrm{d}x+\int\coth^2{x}\operatorname{csch}^2{x}\,\mathrm{d}x \end{align}$$

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$$\begin{align} \int \frac{1}{\sinh^4 x}dx & =16\int\frac{1}{(e^x-e^{-x})^4} dx \end{align}$$ Substitute $u=e^x, du=udx$

$$\begin{align} \int \frac{1}{\sinh^4 x}dx &=16 \int\frac{1}{u} \cdot \frac{1}{(u-1/u)^4} du \\ & = 16\int \frac{u^3}{(u^2-1)^4} du \end{align}$$ Substitute $u^2-1=v, dv=2udu$ $$\int \frac{1}{\sinh^4 x}dx =8\int \frac{v+1}{v^4} dv $$ Now use power law and replace back everything.

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