@Bohemianrelativist Yes. It follows quite logically (if you think about known integrals, the only plausible one that has a 1/(something^2+something else) is the arctan)
If the middle step isn't clear here's the full derivation: $dt/dr=1/\sqrt{-M^2+Q^2}$ and $$(2M^2-Q^2)\int\frac{1}{(r-M)^2-M^2+Q^2}dr=(2M^2-Q^2)\int\frac{1}{(-M^2+Q^2)t^2-M^2+Q^2}\sqrt{-M^2+Q^2}dt=\dfrac{2M^2-Q^2}{\sqrt{-M^2+Q^2}}\text{arctan}(u)$$
since $\text{arctan}'(x)=1/\sqrt(1+x^2)$ this gives $$\dfrac{2M^2-Q^2}{\sqrt{-M^2+Q^2}}\text{arctan}(t)$$$$\dfrac{2M^2-Q^2}{\sqrt{-M^2+Q^2}}\text{arctan}(\sqrt{(r-M)^2/(-M^2+Q^2)})$$
It's a standard constant/second degree polynomial integral, you can complete the square: $$(2M^2-Q^2)\int\frac{1}{r^2-2Mr+Q^2}dr=(2M^2-Q^2)\int\frac{1}{(r-M)^2-M^2+Q^2}dr$$ and then it's an arctan integral if you let $t^2=(r-M)^2/(-M^2+Q^2)$ you'll get $$\dfrac{2M^2-Q^2}{\sqrt{-M^2+Q^2}}\int\frac{1}{1+t^2}dt$$
you could manipulate it to $2Mr-Q^2=M(2r)-Q^2=M(2r-2M+2M)-Q^2=2M(r-M)+2M^2-Q^2$. Then you should split into two fractions, the first will give the logarithm integral, the second will be constant/second degree polynomial with has a formula or can be done quite simply by completing the square and forming an arctan
Well as it's a fraction you could take the derivative of the denominator and see that it's $2(r-M)$ because you want to form an integrand of the type $f'/f$ which integrates to $\ln f$. So to form $2(r-M)$ in the numerator
@cairdcoinheringaahing cuz now the non-uk costs for the top unis (only ones worth applying to abroad, of course) are like over 20 30k and i feel like it's too much of a hassle to try to get the very few grants available (and i wouldn't spend that much on my undergrad)
@cairdcoinheringaahing Going well overall, in terms of school and like social life, but I'm kinda tired :) currently looking for Uni opportunities and I'm kinda frustrated that all the options in the UK are now sort of ruled out for me, but I'll figure it out
