Originally Posted by

**mellowdano** **I had to Complete the perfect square and use the tables of integration to integrate the following:**

**∫ 1 / (√x^2 – 3x + 6) dx**

**What I have done:**

**Completing the perfect square we have**;

x^2 – 3x + 6 = x^2 – 3x + 2.25 + 6 - 2.25 = (x – 1.5)^2 +3.75

**if we set u = x – 1.5, then du = dx, the integral in terms of u is**

∫ 1 / √(u^2 – 2.25) du

**We then have**

∫ 1 / √(u^2 – 2.25) du = ln|u +√u^2 – 2.25|,

**And the answer in terms of x is**

ln|x + 2 +√(u^2 – 2.25)| + C

Have I done this correctly?