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!
 A: $$I=\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \int_0^1\frac{4x-5}{\sqrt{4-(x-1)^2}}dx$$
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?
A: 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 $\tau=-1$.
$$\int\frac{4x-5}{\sqrt{3+2x-x^{2}}}=\mu\int\frac{d(3+2x-x^{2})}{\sqrt{3+2x-x^{2}}}+\tau\int\frac{dx}{\sqrt{3+2x-x^{2}}}$$$$= -2(2\sqrt{(3+2x-x^{2}})+-1\left(\int\frac{d(x-1)}{\sqrt{2^{2}-(x-1)^{2}}}\right) $$$$=-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$
A: 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.
