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!

On solving we will find that it is equal to -$$-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$
Now if you put the appropriate limits I guess you'll get your answer.
First of all write $$4x-5 = \mu \frac{d(3+2x-x^{2})}{dx}+\tau(3+2x-x^{2})$$
We will find that $\mu=-2$ and $\tau=-1$.
$$\int\frac{4x-5}{\sqrt{3+2x-x^{2}}}=\mu\int\frac{d(3+2x-x^{2})}{\sqrt{3+2x-x^{2}}}+\tau\int\frac{dx}{\sqrt{3+2x-x^{2}}}$$$$= -2(2\sqrt{(3+2x-x^{2}})+-1\left(\int\frac{d(x-1)}{\sqrt{2^{2}-(x-1)^{2}}}\right) $$$$=-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$
