Question

Solve $\int \frac{1}{\sqrt{8+3x-x^2}}dx$

Answer 1

$\begin{array}{c}\int \frac{dx}{\sqrt{8+3x-x^2}}\\ =\int \frac{dx}{-\sqrt{x^2-3x-8}}\\ =\int \frac{dx}{-\sqrt{\left(x^2-2.\frac{3}{2}.x+\frac{9}{4}\right)-\frac{9}{4}-8}}\\ =\int \frac{dx}{-\sqrt{{\left(x-\frac{3}{2}\right)}^2-\frac{41}{4}}}\\ =\int \frac{dx}{\sqrt{\frac{41}{4}-{\left(x-\frac{3}{2}\right)}^2}}\\ =\int \frac{dx}{\sqrt{{\left(\frac{\sqrt{41}}{2}\right)}^2-{\left(x-\frac{3}{2}\right)}^2}}\\ ={\sin }^{-1}\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}+C\\ ={\sin }^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C\end{array}$