How do I evaluate this integral in calculus?

Question

∫dx/x^2-6x+9 dx with limits [1,2] I know I should factor out the denominator, but then what?

Answer 1

Given function is ∫dx/(x²-6x+9)

First factorize denominator

     x²-6x+9

     x²-3x-3x+9

     x(x-3)-3(x-3)

     (x-3)(x-3)

     (x-3)²

= ∫dx/(x²-6x+9)

= ∫dx/(x-3)²

u = x-3 ⇒ du = dx

= ∫dx/u²

= -1/u

Substitute u = x-3

= -1/(x-3)

Substitute interval [1,2]

= [-1/(2-3)] -[-1/(1-3)]

= 1(-1/-1)+(-1/-2)

= 1-(1/2)

= 1/2