`int sqrt(5 + 4x - x^2) dx` Evaluate the integral

`intsqrt(5+4x-x^2)dx`

Rewrite the integrand by completing the square: 

`=intsqrt(-(x-2)^2+9)dx`

Now apply the integral substitution,

Let u=x-2,

`=>du=dx`

`=intsqrt(9-u^2)du`

Now using the standard integral:

`intsqrt(a^2-x^2)dx=(xsqrt(a^2-x^2))/2+a^2/2sin^(-1)x/a+C`  

`intsqrt(9-u^2)du=(usqrt(3^2-u^2))/2+3^2/2sin^(-1)u/3`

`=(usqrt(9-u^2))/2+9/2sin^(-1)u/3`

Substitute back u=x-2,

`=((x-2)sqrt(9-(x-2)^2))/2+9/2sin^(-1)((x-2)/3)`

Add a constant C to the solution,

`=((x-2)sqrt(9-(x-2)^2))/2+9/2sin^(-1)((x-2)/3)+C`