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!

$$I=\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \int_0^1\frac{4x-5}{\sqrt{4-(x-1)^2}}dx$$
Let $x-1=2\sin\theta\implies dx=2\cos\theta d\theta$
If $x=0, \sin\theta=\frac12, \theta=\frac\pi6$
If $x=1, \sin\theta=0, \theta=0$
$$\text{So,}I=\int_{\frac\pi6}^0\frac{4(2\sin\theta+1)-5}{2\cos\theta}\cdot2\cos\theta d\theta$$
$$=8\int_{\frac\pi6}^0\sin\theta d\theta-\int_{\frac\pi6}^0\theta d\theta$$
$$=8[-\cos\theta]_{\frac\pi6}^0-\left(0-\frac\pi6\right)$$
Can you take it home from here?
