I am struggling to understand the method of solving the integral of $\sqrt{(3x-x^2-1)}$. What I've done so far is to extend it by itself to create a rational expression an divide it into $\frac{3x}{\sqrt{(3x-x^2-1)}}-\frac{x^2}{\sqrt{(3x-x^2-1)}}-\frac{1}{\sqrt{(3x-x^2-1)}}$. Furthermore, I have solved all terms except the $x^2$-term. I appreciate if anyone could give me a tips or help me with this one as I want to make sure I have understood the concepts correctly. I do have tried integration by parts but could not solve it anyway.
Thanks in advance!
Hint to compute $\int\sqrt{3x-x^2-1}dx$.
we have
$$\sqrt{3x-x^2-1}=\sqrt{\frac{5}{4}-(x-\frac{3}{2})^2}$$
$$=\frac{\sqrt{5}}{2}\sqrt{1-(\frac{2x-3}{\sqrt{5}})^2}$$
then, we put
$$\frac{2x-3}{\sqrt{5}}=\sin(t)$$
you could finish.