How does one go from this: $ \int \frac{l_0(1 − ξ/x_1)}{\sqrt{κ_0^2(1 − x_0/x_1)^2 − l_0^2(1 − ξ/x_1)^2}} dξ$
to this: $ \frac{x_1}{l_0}\sqrt {κ_0^2(1 − x_0/x_1)^2 − l_0^2(1 − ξ/x_1)^2}$
What i've tried to to is integrate via $u$ substitution, with $u = ξ/x_1$ and i was very close to the expression in the second line however i got out a minus sign at the front which isn't correct. Is there a simple way of getting to the second line?
The first thing to recognise here is that most of the integral can be replaced with constants since they aren't in terms of $\xi$. To do this, we'll define $a:=k_0(1-x_0/x_1)$ (this is mostly to make the question easier to work through).
We then make the substitution $u=l_0(1-\xi/x_0)$ which also gives $-\frac{x_0}{l_0}\mathrm{d}u=\mathrm{d}\xi$ and allowing us to transform the integral into $$ \frac{x_0}{l_0}\int\frac{-u}{\sqrt{a^2-u^2}}\mathrm{d}u $$
For this integral, we need to use inspection. If we consider the $\sqrt{a^2-u^2}$, the derivative of this is $$ \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}u}\sqrt{a^2-u^2}&=\frac{1}{2}\frac{1}{\sqrt{a^2-u^2}}\frac{\mathrm{d}}{\mathrm{d}u}\left(a^2-u^2\right)\\ &=\frac{1}{2}\frac{1}{\sqrt{a^2-u^2}}\left(-2u\right)\\ &=\frac{-u}{\sqrt{a^2-u^2}} \end{aligned} $$ This happens to be our integrand, so we can now calculate the integral as follows $$ \begin{aligned} \frac{x_0}{l_0}\int\frac{-u}{\sqrt{a^2-u^2}}\mathrm{d}u&=\frac{x_0}{l_0}\int\frac{\mathrm{d}}{\mathrm{d}u}\sqrt{a^2-u^2}\mathrm{d}u\\ &=\frac{x_0}{l_0}\left[\sqrt{a^2-u^2}+c\right] \end{aligned} $$
When you replace the substitutions you'll arrive at your solution.
If you're wondering where the $\sqrt{a^2-u^2}$ came from, you'll notice that in the original integrand, the numerator is the derivative of the inside of the square root. When you have the integral of the product of two functions where one is related to the derivative of the other, it's always worth trying to differentiate that function and see what you get. Sometimes it works out quite nice for you