# Integral of 1/polynomial (order 2)

1. Apr 25, 2007

1. The problem statement, all variables and given/known data

$$\int{ \frac{dx}{Ax^2 + Bx + C}$$

3. The attempt at a solution

So I can't think of any immediately obvious substitutions. What I've tried is completing the square in the denominator so that the integrand becomes

$$\frac{1}{(\sqrt{A}x + \frac{B}{2\sqrt{A}})^2 - (\frac{B^2}{4A} - C)}$$

I guess then I could treat it as a difference of two squares, then use partial fractions? That's going to be a lot of work though, and I was wondering if it will even work or if there's a better way of doing it.

Last edited: Apr 25, 2007
2. Apr 25, 2007

### Glass

I would try partial fractions over the complex numbers. BTW, I've never actually done it, but see no reason why it couldn't be done. On the other hand, you may as well give it a shot while you wait for someone more qualified to answer.

Last edited: Apr 25, 2007
3. Apr 25, 2007

### mezarashi

This is an inverse trigonometric integral, meaning the solution involves an inverse tan.

The strategy is to complete the square of the denominator so that you have something of the form:

$$\int{ \frac{dx}{(x+\frac{b}{2a})^2 + \frac{c}{a} - \frac{b^2}{4a^2}}$$

With that, you can make the substitution of $$u = x + \frac{b}{2a}$$, followed by the substitution $$u = d cot \theta$$. Through the trigonometric substition $$cot^2 \theta + 1 = tan^2 \theta$$ you will then have a simple $$tan^2 \theta$$ in the denominator, which can then once again be substituted into a $$\frac{1}{v^2}$$ problem.

Hope that helps.

4. Apr 25, 2007