Answer
$$\sqrt 3 \ln | \sinh (\dfrac{\theta}{\sqrt 3})| +C= \sqrt 3 \ln |e^{\theta/\sqrt{3}}-e^{-\theta/\sqrt{3}}| +C$$
Work Step by Step
Given: $\int \coth (\dfrac{\theta}{\sqrt 3}) d\theta$
Plug in $\dfrac{\theta}{\sqrt 3}=t$ and $d \theta =\sqrt 3 dt$
Thus, $\int \coth (\dfrac{\theta}{\sqrt 3}) d\theta=\sqrt 3 \int \coth t dt=\sqrt 3 \int \dfrac{\cosh t}{\sinh t} dt$
Now, plug in $\sinh t =u \implies \cosh t dt =du$
Thus, we have:
$=\sqrt 3 \int \dfrac{du}{u}$
or, $= \sqrt 3 \ln |u| +C$
$= \sqrt 3 \ln | \sinh (\dfrac{\theta}{\sqrt 3})| +C$
$= \sqrt 3 \ln | \frac{e^{\theta/\sqrt{3}}-e^{-\theta/\sqrt{3}}}{2}| +C$
$= \sqrt 3 \ln |e^{\theta/\sqrt{3}}-e^{-\theta/\sqrt{3}}| +C$
