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I've been following a derivation of relativistic kinetic energy. I've seen other ways to get the end result but I'm interested in finding out where I've gone wrong here: I'm struggling with integrating by parts.

The author goes from $$\frac{\textrm{d}T}{\textrm{d}t}=\frac{m}{\sqrt{(1-u^2/c^2)^3}}u\frac{\textrm{d}u}{\textrm{d}t}$$to$$\frac{\textrm{d}T}{dt}=\frac{\textrm{d}}{\textrm{d}t}\left[\frac{mc^2}{\sqrt{(1-u^2/c^2)}}\right].$$

Are they are using a chain rule like$$

\frac{\textrm{d}T}{\textrm{d}t}=\frac{\textrm{d}T}{\textrm{d}u}\frac{\textrm{d}u}{\textrm{d}t}?$$

If so then I need to find ##\frac{\textrm{d}T}{\textrm{d}u}.##

From the first equation I have$$\textrm{d}T = \frac{m}{\sqrt{(1-u^2/c^2)^3}}u\textrm{d}u.$$ If integrating by parts is $$\int A\frac{\textrm{d}B}{\textrm{d}u}\textrm{d}u = AB-\int\frac{\textrm{d}A}{\textrm{d}u}B\textrm{d}u$$ then I choose ##A = \sqrt{(1-u^2/c^2)^3}## and ##\frac{\textrm{d}B}{\textrm{d}u}= u.## So ##B = u^2/2## and from the chain rule$$\frac{\textrm{d}A}{\textrm{d}u}=\frac{3u}{c^2\sqrt{\left(1-\frac{u^2}{c^2}\right)^5}},$$ but this gets me right back to the start with an integral of the form $$\int u f(u) \textrm{d}u.$$ Choosing A and B' the other way round in the integration by parts doesn't give me an integral I can solve by inspection to find B.

Where have I gone wrong?

Thanks

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# I Integration by parts

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