Integration of rational function with quadratic denominator

Outline of method

Reduction to the case where the numerator is constant or linear and the denominator is monic

Fill this in later

Case that the denominator has distinct linear factors

UPSHOT: The antiderivative in this case is expressible as a linear combination with constant coefficients of the natural logarithms of the absolute values of the linear factors.

This falls under the general case of integration of rational function whose denominator has distinct linear factors

The integration formula is:

$\int \frac{Ax+B}{(x-{\alpha }_1)(x-{\alpha }_2)}dx=\frac{A{\alpha }_1+B}{{\alpha }_1-{\alpha }_2}\ln |x-{\alpha }_1|+\frac{A{\alpha }_2+B}{{\alpha }_2-{\alpha }_1}\ln |x-{\alpha }_2|+C$

Note that ${\alpha }_1$ and ${\alpha }_2$ can be determined from the quadratic formula for the roots of a quadratic polynomial. Specifically, if the polynomial in the denominator is $x^2+px+q$, we have:

${\alpha }_1=\frac{-p+\sqrt{p^2-4q}}{2},{\alpha }_2=\frac{-p-\sqrt{p^2-4q}}{2}$

Here are the details of how the formula is obtained:

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We want to write:

$\frac{Ax+B}{(x-{\alpha }_1)(x-{\alpha }_2)}=\frac{c_1}{x-{\alpha }_1}+\frac{c_2}{x-{\alpha }_2},c_1,c_2\in \mathbb{R}$

First, note that this is possible, because the denominator has higher degree than the numerator.

Multiplying both sides by the denominator, we get:

$Ax+B=c_1(x-{\alpha }_2)+c_2(x-{\alpha }_1)$

Plugging $x={\alpha }_1$ in the above, we get:

$A{\alpha }_1+B=c_1({\alpha }_1-{\alpha }_2)⟹c_1=\frac{A{\alpha }_1+B}{{\alpha }_1-{\alpha }_2}$

Similarly, plugging $x={\alpha }_2$ instead gives:

$A{\alpha }_2+B=c_2({\alpha }_2-{\alpha }_1)⟹c_2=\frac{A{\alpha }_2+B}{{\alpha }_2-{\alpha }_1}$

Plugging the values of $c_1,c_2$ thus obtained into the original expression, we get:

$\frac{Ax+B}{(x-{\alpha }_1)(x-{\alpha }_2)}=\frac{A{\alpha }_1+B}{{\alpha }_1-{\alpha }_2}\cdot \frac{1}{x-{\alpha }_1}+\frac{A{\alpha }_2+B}{{\alpha }_2-{\alpha }_1}\frac{1}{x-{\alpha }_2}$

Integrating both sides, we obtain:

$\int \frac{Ax+B}{(x-{\alpha }_1)(x-{\alpha }_2)}dx=\int \frac{A{\alpha }_1+B}{{\alpha }_1-{\alpha }_2}\cdot \frac{1}{x-{\alpha }_1}dx+\int \frac{A{\alpha }_2+B}{{\alpha }_2-{\alpha }_1}\frac{1}{x-{\alpha }_2}dx=\frac{A{\alpha }_1+B}{{\alpha }_1-{\alpha }_2}\ln |x-{\alpha }_1|+\frac{A{\alpha }_2+B}{{\alpha }_2-{\alpha }_1}\ln |x-{\alpha }_2|+C$

Case that the denominator has repeated linear factors

UPSHOT: The antiderivative in this case is a constant divided by the linear factor plus a constant times the natural logarithm of the linear factor.

Fill this in later

Case that the denominator has negative discriminant

UPSHOT: The antiderivative in this case is a constant times an arc tangent function plus a constant times the natural logarithm of the absolute value of the quadratic.

We have:

Failed to parse (unknown function "\gammma"): {\displaystyle \int \frac{Ax + B}{(x - \beta)^2 + \gamma^2} \, dx = \frac{A}{2}\ln((x - \beta)^2 + \gammma^2) + \frac{B + A\beta}{\gamma} \arctan\left(\frac{x - \beta}{\gamma}\right) + C}

Given a denominator in the form $x^2+px+q$, it can be rewritten as $(x-\beta {)}^2+{\gamma }^2$ where:

$\beta =\frac{-p}{2},\gamma =\sqrt{q-(p^2/4)}$