TL;DR

I don't understand how Tipler/Mosca famous book accomplishes Relativistic Energy

As in classic mechanics, we will define kinetic energy as the work done by the net force in accelerating a particle from rest to some final velocity ##u_f##. Considering one dimension only, we have

$$K=\displaystyle\int_{u=0}^{u=u_f}F_{net}\,ds=\displaystyle\int_{u=0}^{u-u_f}\cfrac{dp}{dt}\,ds=\displaystyle\int_{u=0}^{u-u_f}u\,dp=\displaystyle\int_{u=0}^{u-u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )\qquad{39-21}$$

where we have used ##u=ds/dt##. It is left as a problem for you to show that

$$d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

(I needed a quick browsing to understand it)

$$K=\displaystyle\int_{u=0}^{u=u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

$$=mc^2\Bigg (\cfrac{1}{\sqrt{1-(u^2/c^2)}}-1\Bigg )$$

This last step is what I don't understand.

Attempt

$$(1-u^2/c^2)=t\Rightarrow{dt=-2u/c^2}$$

But it doesn't work: I can't get rid of ##u##. I neither know how to deal with ##c##

$$K=\cfrac{mc^2}{\sqrt{1-(u^2/c^2)}}-mc^2\qquad{39-22}$$

REST ENERGY

$$E_0=mc^2\qquad{39-23}$$

RELATIVISTIC ENERGY

$$E=K+mc^2=\cfrac{mc^2}{\sqrt{1-(u^2/c^2)}}\qquad{39-24}$$

$$\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du=m\displaystyle\int_0^{u_f}\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

$$1-\cfrac{u^2}{c^2}=t\Rightarrow{-\cfrac{2u}{c^2}}\,du=dt\Rightarrow{u\,du=-\cfrac{c^2}{2}\,dt}$$

$$m\Big (-\frac{c^2}{2}\Big )\displaystyle\int_0^{u_f}\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,dt$$

$$-m\cfrac{c^2}{2}\displaystyle\int_0^{u_f}t^{-3/2}\,dt=-m\frac{c^2}{2}\cdot\cfrac{t^{-1/2}}{-1/2}\Bigg |_0^{u_f}$$

$$mc^2\Bigg (\cfrac{1}{\sqrt{1-\cfrac{u_f^2}{c^2}}}-1\Bigg )$$

Last step is this way

$$mc^2t^{-1/2}\bigg |_{1}^{1-\frac{u_f^2}{c^2}}=mc^2\left (\frac{1}{\sqrt{1-u_f^2/c^2}}-1\right )$$