Integrating with Partial Fractions - Irreducable

[kathrynag]

Homework Statement

I need to integrate (2x-5)/(x^2+5x+11)

Homework Equations

The Attempt at a Solution

My problem is just finding a formula for an irreducable quadratic. I know if the denominator was x(x^2+1), I would use A/x+(Bx+C)/(x^2+1). I just don't know the formula in this case and I feel like I could solve it if I had a formula.

 

[snipez90]

Rewrite the integrand as

[tex]\frac{2x - 5 + 10 - 10}{x^2 + 5x + 11} = \frac{2x + 5}{x^2 + 5x + 11} - \frac{10}{x^2 + 5x + 11}.[/tex]

In the new expression, we can integrate the first by noting that the numerator is the derivative of the denominator. For the second, complete the square and substitute.

 

[kathrynag]

Rewrite the integrand as

[tex]\frac{2x - 5 + 10 - 10}{x^2 + 5x + 11} = \frac{2x + 5}{x^2 + 5x + 11} - \frac{10}{x^2 + 5x + 11}.[/tex]

In the new expression, we can integrate the first by noting that the numerator is the derivative of the denominator. For the second, complete the square and substitute.

Ok, I sort of understand that, but let's say the numerator was something like 3x-7. Is there something different to do in this case?

 

[snipez90]

Not really, the basic idea is to ensure that we have some expression where the derivative of the quadratic in the denominator is the numerator.

If 3x - 7 was in the numerator, then 3x - 7 = (3/2)(2x - 14/3). Hence,

[tex]\frac{3x - 7}{x^2 + 5x + 11} = \frac{3}{2}\cdot\frac{2x - 14/3 + 29/3 - 29/3}{x^2 + 5x + 11} = \frac{3}{2}\left(\frac{2x + 5}{x^2 + 5x + 11} - \frac{29}{3}\cdot\frac{1}{x^2 + 5x + 11}\right)[/tex]

 

[Gregg]

That's a good way to do it, remembering that:

[tex]\int \frac{f'(x)}{f(x)} dx = \log_e (f(x))[/tex]

and that

[tex]\int \frac{dx}{x^2+a^2} = \frac{\tan^{-1} (\frac{x}{a}) }{a} [/tex]

Seeing that the integral can be split in the fashion described is a good start and maybe it's the first step you should look to take when seeing an integrand that is a quotient of two polynomials the numerator being of one less degree than the denominator.

 

[kathrynag]

Thanks! I figured it out!

 

[Gregg]

I think this would be general for problems like these?

[tex] \int \frac{Ax+B}{ax^2+bx+c} dx = \frac{A}{2a}\int \frac{2a}{A}.\frac{Ax+B}{ax^2+bx+c} = \frac{A}{2a}\left( \int\frac{2ax+b}{ax^2+bx+c}+\int{\frac{\frac{2aB}{A}-b}{ax^2+bx+c}\right)[/tex]

[tex] \int \frac{\frac{2aB}{A}-b}{ax^2+bx+c} = \frac{\frac{2aB}{A}-b}{a\sqrt{c-\frac{b^2}{4a^2}}}\tan^{-1}\left(\frac{x+\frac{b}{2a}}{\sqrt{c-\frac{b^2}{4a^2}}}\right)[/tex]

So

[tex]\int \frac{Ax+B}{ax^2+bx+c} dx = \frac{A}{2a}\left(log_e(ax^2+bx+c) + \frac{\frac{2aB}{A}-b}{a\sqrt{c-\frac{b^2}{4a^2}}}\tan^{-1}\left(\frac{x+\frac{b}{2a}}{\sqrt{c-\frac{b^2}{4a^2}}}\right)\right)[/tex]

 

[kathrynag]

Not really, the basic idea is to ensure that we have some expression where the derivative of the quadratic in the denominator is the numerator.

If 3x - 7 was in the numerator, then 3x - 7 = (3/2)(2x - 14/3). Hence,

[tex]\frac{3x - 7}{x^2 + 5x + 11} = \frac{3}{2}\cdot\frac{2x - 14/3 + 29/3 - 29/3}{x^2 + 5x + 11} = \frac{3}{2}\left(\frac{2x + 5}{x^2 + 5x + 11} - \frac{29}{3}\cdot\frac{1}{x^2 + 5x + 11}\right)[/tex]

Where does the 29/3 come from? I see that we want 2x+5 in the numerator. So we would have 3x-7+12-12, right? From there, I get (3/2)(2x-14/3+24/3-24/3). This will give us what we want. I'm just trying to see where 29/3 came from instead of 24/3?

 

[kathrynag]

I think I figured it out. It's because of the 3/2 factored out?