Strange solution for the integral

Question

Reading some examples in my theory book, I met strange solution and I can't figure out how did they got it. Here it is $$\int \frac{v \, dv}{\sqrt{v^4 +g^2r^2}}=\frac{1}{2}\ln{ \frac{v^2+\sqrt{v^4+g^2r^2}}{gr}}+C,$$ where $g,r$ are constants. I can't understand why they have $gr$ in the denominator. Is it a misprint, or it has some meaning which I miss?

Answer 1

This example looks like it comes from physics - the $gr$ in the denominator of the log is there to make the argument of the log a dimensionless number. note that it can be absorbed into the definition of $C$ ( provided that $gr$ is positive )

Answer 2

Since $g$ and $r$ constants, we can write \begin{align} \frac{1}{2}\ln{ \frac{v^2+\sqrt{v^4+g^2r^2}}{gr}}+C &=\frac{1}{2}\ln{\left(v^2+\sqrt{v^4+g^2r^2} \ \right)}- \frac{1}{2}\ln(gr)+C \\ &=\frac{1}{2}\ln{\left(v^2+\sqrt{v^4+g^2r^2} \ \right)}+D. \end{align} Anyhow, they are equivalent.

Answer 3

\begin{align} & \int \frac{v \, dv}{\sqrt{v^4 +g^2r^2}} = \int \frac{\frac 1 2\,du}{\sqrt{u^2+g^2 r^2}} = \frac 1 2 \int \frac{du/gr}{\sqrt{\left(\frac u{gr}\right)^2 + 1}} = \underbrace{\frac 1 2 \int \frac{\sec^2\theta\,d\theta}{\sqrt{\tan^2\theta + 1}}}_{\text{No $gr$ appears here.}} \\[12pt] = {} & \overbrace{\frac 1 2 \int \sec\theta\,d\theta}^{\text{No $gr$ appears here.}} = \frac 1 2 \ln(\tan\theta+\sec\theta) + \text{constant} \\[12pt] = {} & \frac 1 2 \ln((u/gr) + \sqrt{(u/gr)^2 + 1}) + \text{constant} \\[12pt] = {} & \frac 1 2 \ln \frac{u + \sqrt{u^2+ g^2r^2}}{gr} + \text{constant} \\[12pt] = {} & \frac 1 2 \ln(u + \sqrt{u^2 + g^2r^2})\, \underbrace{{}-\ln(gr) + \text{constant}}_{\text{This is a constant.}} \end{align}

You don't need $gr$ in the answer; it is equally correct to write it without that, however: it's correct either way, since the answer with $gr$ and that without it differ from each other by a constant, which is $\ln(1/gr)$.

But the argument above shows how it can get there.

Remember that "constant" in this case means not changing as the variable $v$ changes.

Answer 4

https://en.wikipedia.org/wiki/Hyperbolic_function#Standard_integrals

Substitute $u=v^2$, that integral can be written as $\sinh^{-1}$