Integrating a school homework question. Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found.
Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$
Thank you in advance!

Note first that the derivative of the quadratic in the denominator is $-2x+2$. It would be great if the numerator were $4x-4$, because then we could set $u=3+2x-x^2$, $du=(-2x+2)dx$, and write the indefinite integral as
$$\int\frac{-2}{\sqrt u}du=-2\int u^{-1/2}du\;.$$
Unfortunately, the numerator is actually $4x-5=-2(-2x+2)-1$. The trick is to split the integral in two:
$$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx=\int_0^1\frac{4x-4}{\sqrt{3+2x-x^2}}dx-\int_0^1\frac{dx}{\sqrt{3+2x-x^2}}\;.$$
We’ve already sorted out how to deal with the first of these integrals, so that leaves only the second. Complete the square: 
$$3+2x-x^2=-(x^2-2x-3)=-\big((x-1)^2-4\big)=2^2-(x-1)^2\;,$$
so
$$\int_0^1\frac{dx}{\sqrt{3+2x-x^2}}=\int_0^1\frac{dx}{\sqrt{2^2-(x-1)^2}}\;,$$
which can be handled with a standard trig substitution.
