Question

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

Answer 1

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

$=\int \frac{dx}{\sqrt{-(-8-3x+x^2)}}$

$=\int \frac{dx}{\sqrt{-(x^2-3x-8)}}$

$=\int \frac{dx}{\sqrt{-(x^2-2\times \frac{3}{2}x+\frac{9}{4}-\frac{9}{4}-8)}}$

$=\int \frac{dx}{\sqrt{-({(x-\frac{3}{2})}^2-\frac{9+32}{4})}}$

$=\int \frac{dx}{\sqrt{-({(x-\frac{3}{2})}^2-\frac{41}{4})}}$

$=\int \frac{dx}{\sqrt{({\left(\frac{\sqrt{41}}{2}\right)}^2-{(x-\frac{3}{2})}^2)}}$

$={\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+c$

$={\sin }^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+c$