Derivation of relativistic E_kin

[tomwilliam2]

In my textbook there is a derivation of relativistic kinetic energy starting from an integral of the force applied over the distance required to take the particle's speed from 0 to v.
There's one stage of the derivation I don't understand on mathematical grounds, which is going from:

$$E_k = \int_0^{v} u\ d\left( \frac{mu}{\sqrt{1-(u^2/c^2)}}\right )$$
To the next line, which is:
$$E_k = \left [ \frac{mu^2}{\sqrt{1-(u^2/c^2)}}\right]_0^v - \int_0^v \frac{mu}{\sqrt{1-(u^2/c^2)}}\ du$$
I guess they have used integration by parts, but how do you get the change of integrating variable? I only really know the rule $\int f'g dx = fg - \int fg' dx$ and can't seem to make it fit here.
Thanks in advance

 

[Simon Bridge]

What was the original variable being integrated over?
What was substituted in it's place?

 

[phyzguy]

Take:
$$f' = d (\frac{mu}{\sqrt{1-(u^2/c^2)}}), g = u$$

 

[tomwilliam2]

Ah, thanks phyzguy...the penny has dropped.

 

[sardar]

I can provide an explanation for this derivation. The integration by parts technique is indeed used to derive the expression for relativistic kinetic energy. Let's break down the steps to understand it better.

First, we start with the definition of work, which is given by the integral of force over distance:

$$W = \int_0^{x} F\ dx$$

In this case, we are considering the work done to accelerate a particle from rest to a velocity of v. So, the distance x in the integral is the distance required to achieve this velocity.

Next, we use the definition of force, which is given by Newton's second law:

$$F = \frac{d}{dt} (mu)$$

Substituting this into the work equation, we get:

$$W = \int_0^{x} \frac{d}{dt} (mu)\ dx$$

Now, we can use the chain rule to rewrite this as:

$$W = \int_0^{x} \frac{du}{dt} \frac{d}{du} (mu)\ dx$$

Since the velocity is changing with time, we can write:

$$\frac{du}{dt} = \frac{du}{dx} \frac{dx}{dt} = v \frac{du}{dx}$$

Substituting this into the integral, we get:

$$W = \int_0^{x} v \frac{du}{dx} \frac{d}{du} (mu)\ dx$$

Now, we can use the chain rule again to rewrite the term $\frac{d}{du} (mu)$ as:

$$\frac{d}{du} (mu) = m \frac{d}{du} (u) + u \frac{d}{du} (m) = m + u \frac{dm}{du}$$

Substituting this back into the integral, we get:

$$W = \int_0^{x} v \frac{du}{dx} (m + u \frac{dm}{du})\ dx$$

Next, we can rearrange this equation to get:

$$W = \int_0^{x} m \frac{du}{dx} v\ dx + \int_0^{x} u \frac{du}{dx} \frac{dm}{du}\ dx$$

Now, the first term