Evaluate: `intsqrt(1+3x-x^2)dx`

Explanation

`" Let I "=int sqrt(1+3x-x^(2))dx`
`=int sqrt(-(x^(2)-3x-1))dx`
`= int sqrt(-[x^(2)-3x+((3)/(2))^(2)-((3)/(2))^(2)])dx`
`=int sqrt(-[(x-(3)/(2))^(2)-(9)/(2)-1])dx`
`=int sqrt(-[(x-(3)/(2))^(2)-(9+4)/(4)])+dx`
`= int sqrt(-[(x-(3)/(2))^(2)-(13)/(4)])dx`
`=int sqrt(((sqrt(13))^(2)/(2))^(2)-(x-(3)/(2))^(2))dx`
`=(x-(3)/(2))/(2) sqrt(1+3x-x^(2))`
`+(13)/(4xx2) sin^(-1) ((x-(3)/(2))/((sqrt(13))/(2)))`
`rArr I=(2x-3)/(4) sqrt(1+3x-x^(2))`
`+(13)/(8)sin^(-1) ((2x-3)/(sqrt(13)))+C`