int 1/(x^2-4x+9) dx Find the indefinite integral

int1/(x^2-4x+9)dx
Let's complete the square for the denominator of the integral as:
(x^2-4x+9)=(x-2)^2+5  
(x-2)^2+(sqrt(5))^2
int1/(x^2-4x+9)dx=int1/((x-2)^2+(sqrt(5))^2)dx
Let's apply the integral substitution,
substitute u=x-2
du=1dx
=int1/(u^2+(sqrt(5))^2)du
Now use the standard integral :int1/(x^2+a^2)=1/aarctan(x/a)
=1/sqrt(5)arctan(u/sqrt(5))
substitute back u=(x-2) and add a constant C to the solution,
=1/sqrt(5)arctan((x-2)/sqrt(5))+C