How to integrate $\int \frac{1}{g-kv^2}~dv$ How to integrate $$\int \frac{1}{g-kv^2}~dv$$ where $v$ is velocity and $g,k$ are known constants.
I have been stuck for a while on this and have now turned to SE for help any suggestions?
 A: I'm assuming $g, k > 0$. Let $\omega^2 = g/k$. Then 
\begin{align}\int \frac{1}{g - kv^2}\, dv &= \frac{1}{k}\int \frac{1}{\omega^2 - v^2}\, dv\\
& = \frac{1}{2\omega k}\int \left(\frac{1}{\omega - v} + \frac{1}{\omega + v}\right)\, dv\\
& = \frac{1}{2\omega k}(-\log|\omega - v| + \log|\omega + v|) + C\\
& = \frac{1}{2\omega k}\log \left|\frac{\omega + v}{\omega - v}\right| + C\\
& = \frac{1}{2\sqrt{gk}}\log\left|\frac{\sqrt{g/k} + v}{\sqrt{g/k} - v}\right| + C.
\end{align}
A: Kobe provided an excellent solution using partial fraction expansions.  So, here, I'll use a different approach - hyperbolic trigonometric substitution.
To that end, let $I$ be the indefinite integral of interest, namely $I=\int \frac{dv}{g-kv^2}$.  Now, let's make the substitution $v=\sqrt{\frac{g}{k}}\tanh x$.  Then, we have $dv=\sqrt{\frac{g}{k}}  \text{sech}^2 x dx$ and 
$$\begin{align}
I & = \int \frac{dv}{g-kv^2} \\
&= \int \frac{\sqrt{\frac{g}{k}}  \text{sech}^2 x \; dx}{g\left(1-\tanh^2x\right)} \\
& =\frac{1}{\sqrt{gk}} x +C\\
& =\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}v\right) +C
\end{align}$$
which is an acceptable final result.  If one prefers, one may use identity $\tanh^{-1} x = \log (\sqrt{\frac{1+x}{1-x}})$ to yield an alternative expression for $I$ as 
$$\begin{align}
I & =\frac{1}{\sqrt{gk}}\log\left(\sqrt{\frac{\sqrt{\frac{g}{k}}+v}{\sqrt{\frac{g}{k}}-v}}\right) +C \\
& =\frac{1}{2\sqrt{gk}} \log \left( \frac{ \sqrt{\frac{g}{k}}+v }{ \sqrt{\frac{g}{k}}-v}\right) +C
\end{align}$$
which recovers the aforementioned result!
