Integral of 1/polynomial (order 2)

[Flux = Rad]

Homework Statement

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

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.

 

[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.

 

[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.

 

[Flux = Rad]

Ahhh yeah. Perfect, thanks for that.

 

[mezarashi]

Okay, you might not want to take my steps in the previous post word for word there. I jumped the gun a bit with the trigonometric substitutions. They require the chain rule, but somehow it cancels each other out in the end. So just work your way through. >.<