How do I compute: $$ \int_0^z (\sqrt{R^2 + x^2} - x)\,dx. $$ This document I am trying to follow has a similar integral to the one I have:
$$ \int (\sqrt{R^2 + x^2} - x) \, dx \\ =x\sqrt{R^2+x^2} + R^2\ln \left(x + \sqrt{R^2 + x^2}\right) + x_0. $$ So I have tried a few things but have gotten no where. Does anyone know how to compute this kinda integral?
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You already have the form of the indefinite integral. So you only need to take the difference.
$$\int_0^z(\sqrt{R^2+x^2}-x){\mathrm d}x=\left[x\sqrt{R^2+x^2}+R^2\ln\left.\left(x+\sqrt{R^2+x^2}\right)+x_0\right]\right|_{x=0}^{z}=z\sqrt{R^2+z^2}+R^2\ln\left(z+\sqrt{R^2+z^2}\right)+x_0-(\ln|R|+x_0)=z\sqrt{R^2+z^2}+R^2\ln\left(z+\sqrt{R^2+z^2}\right)-\ln|R|.$$
If your question is about how to evaluate the indefinite integral, see N.S.'s answer.
$$\int_0^z (\sqrt{R^2 + x^2} - x) \, dx=\int_0^z \sqrt{R^2 + x^2} \, dx-\int_0^z x \, dx.$$
The first one is a standard $x=R \tan(u)$ substitution, and the second is easy.