$$\begin{align} \int\coth^4xdx &= \int \coth^2x\cdot \coth^2x dx \\ &=\int\coth^2x\cdot (\operatorname{csch}^2x-1)dx \\ &=\int \coth^2x\cdot \operatorname{csch}^2x\cdot dx-\int\coth^2x\cdot dx\\ &=-\int u^2 du - \int \coth^2xdx \\ &=-\frac{\coth^3}{3}-\int (\operatorname{csch}^2x-1)dx \\ &=-\frac{\coth^3}{3}+\coth^x+x+C \end{align}$$
My book tells me the right answer is $-\frac{\coth^3}{3}-\coth^x+x+C$
I know this looks like I am nitpicking but I want to make sure I got the identities right. I double checked but can't find the mistake. Where did it go wrong?
As an4s says, $\coth^2x=\frac{1}{\tanh^2x}$ and $\operatorname{csch^2}x=\frac{1}{\sinh^2 x}$, $$1+\operatorname{csch}^2x=\frac{1+\sinh^2 x}{\sinh^2x}=\frac{\cosh^2x-\sinh^2x+\sinh^x}{\sinh^2x}=\frac{1}{\tanh^2x}$$ where $\cosh^2x-\sinh^2x=1$..
Therefore, $$\begin{align} \int\coth^4xdx &= \int \coth^2x\cdot \coth^2x dx \\ &=\int\coth^2x\cdot (\operatorname{csch}^2x+1)dx \\ &=\int \coth^2x\cdot \operatorname{csch}^2x\cdot dx+\int\coth^2x\cdot dx\\ &=-\int u^2 du + \int \coth^2xdx \\ &=-\frac{\coth^3}{3}+\int (\operatorname{csch}^2x+1)dx \\ &=-\frac{\coth^3}{3}-\coth^x+x+C \end{align}$$