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?