How can I integrate the relativistic mass equation using an elegant method?

[Semo727]

Hello!

I would like to count (see the way how to count) this integral
\int_0^v \frac{1}{(1-v^2)^{3/2}} \,dv
It should be
\frac{v}{\sqrt{1-v^2}}.
I have managed to count it (I have just derived the result, and followed steps in reversed order),
but this method was a little bit clumsy, I think.

It looked like this:
\int\frac{1}{(1-v^2)^{3/2}} \,dv=\int\frac{1-v^2+v^2}{(1-v^2)^{3/2}} \,dv=\int\left[\frac{1}{\sqrt{1-v^2}}+\frac{v^2}{(1-v^2)^{3/2}}\right]\,dv=
=\frac{v}{\sqrt{1-v^2}}\ +\ C\ -\ \int\frac{v^2}{(1-v^2)^{3/2}}\,dv\ +\ \int\frac{v^2}{(1-v^2)^{3/2}}\,dv=\frac{v}{\sqrt{1-v^2}}\ +\ C

so

\int_0^v \frac{1}{(1-v^2)^{3/2}} \,dv=\frac{v}{\sqrt{1-v^2}}\ +\ C\ -\ C\ =\frac{v}{\sqrt{1-v^2}}

I would really appreciate if you write here some more elegant way (if exists) to count that integral.
I need to integrate it to get relativistic mass m_v=\frac{m_0}{\sqrt{1-v^2/c^2}}.

 

[quasar987]

I for one admire the way you did it.

 

[Semo727]

OK, thanks. I just thought that there is some much shorter way I don't know about becouse I don't know much about integrating methods. (I know just per partes and substitution)

 

[StatusX]

One way to check if your way of integrating is the fastest is to differentiate your answer and simplify to get back to the integrand. Then compare this process to the way you integrated. In this case, to differentiate, you have to use the product rule, then multiply the first term by (1-v²)/(1-v²), and then add the terms. This is exactly the reverse of how you integrated, which was to separate the integrand into the sum of two terms, multiply the first one by (1-v²)/(1-v²), and then use integration by parts (which you call counting), which is the integral equivalent to the product rule. This would suggest your way is the best.