Integration of rational function with quadratic denominator

Outline of method

Reduction to the case where the numerator has smaller degree than the denominator

For further information, refer: converting a rational function from improper fraction to mixed fraction form

To get to the situation where the numerator has smaller degree than the denominator, we perform a Euclidean division and hence rewrite the rational function as the sum of a polynomial and a rational function that is in proper fraction form, i.e., the numerator has smaller degree than the denominator. The polynomial summand is integrated termwise using the integration rule for power functions. We are thus reduced to handling the proper fraction. But remember to add back the antiderivative for the polynomial to your final answer!

Reduction to the monic denominator case

If the leading coefficient (i.e., the coefficient on the highest degree term) in the denominator is not 1, the leading coefficient can be pulled out as a constant factor from the denominator and hence out of the integration. We can thus carry out an integration with a leading coefficient of 1 for the denominator polynomial (such a polynomial is termed a monic polynomial). But remember to keep that constant on the outside and multiply it to get your final answer!

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:

$\alpha _{1}={\frac {-p+{\sqrt {p^{2}-4q}}}{2}},\qquad \alpha _{2}={\frac {-p-{\sqrt {p^{2}-4q}}}{2}}$

Here are the details of how the formula is obtained:

[SHOW MORE]

We want to write:

${\frac {Ax+B}{(x-\alpha _{1})(x-\alpha _{2})}}={\frac {c_{1}}{x-\alpha _{1}}}+{\frac {c_{2}}{x-\alpha _{2}}},\qquad c_{1},c_{2}\in \mathbb {R}$

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:

$A\alpha _{1}+B=c_{1}(\alpha _{1}-\alpha _{2})\qquad \implies c_{1}={\frac {A\alpha _{1}+B}{\alpha _{1}-\alpha _{2}}}$

Similarly, plugging $x=\alpha _{2}$ instead gives:

$A\alpha _{2}+B=c_{2}(\alpha _{2}-\alpha _{1})\qquad \implies c_{2}={\frac {A\alpha _{2}+B}{\alpha _{2}-\alpha _{1}}}$

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.

The integration formula is:

$\int {\frac {Ax+B}{(x-\sigma )^{2}}}\,dx=A\ln |x-\sigma |-{\frac {A\sigma +B}{x-\sigma }}+C$

If the denominator is of the form $x^{2}+px+q$, then this case arises iff $p^{2}=4q$ and we have $\sigma =-p/2$.

Here's how the formula is obtained: [SHOW MORE]

We rewrite the integrand as:

${\frac {A(x-\sigma )+B+A\sigma }{(x-\sigma )^{2}}}$

It now gets split as:

${\frac {A}{x-\sigma }}+{\frac {A\sigma +B}{(x-\sigma )^{2}}}$

We now integrate term-wise.

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:

$\int {\frac {Ax+B}{(x-\beta )^{2}+\gamma ^{2}}}\,dx={\frac {A}{2}}\ln((x-\beta )^{2}+\gamma ^{2})+{\frac {B+A\beta }{\gamma }}\arctan \left({\frac {x-\beta }{\gamma }}\right)+C$

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

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

Here's how the formula is obtained: [SHOW MORE]

We want to rewrite the numerator as a linear combination of 1 and the derivative of the denominator:

$Ax+B=c_{1}.(2(x-\beta ))+c_{2}.1$

Simplifying, we get:

$2c_{1}=A,\qquad -2c_{1}\beta +c_{2}=B$

We thus get $c_{1}=A/2,c_{2}=B+A\beta$.

Plugging these back in, we get:

$\int {\frac {Ax+B}{(x-\beta )^{2}+\gamma ^{2}}}\,dx=c_{1}\int {\frac {2(x-\beta )\,dx}{(x-\beta )^{2}+\gamma ^{2}}}+c_{2}\int {\frac {dx}{(x-\beta )^{2}+\gamma ^{2}}}={\frac {A}{2}}\int {\frac {2(x-\beta )\,dx}{(x-\beta )^{2}+\gamma ^{2}}}+(B+A\beta )\int {\frac {dx}{(x-\beta )^{2}+\gamma ^{2}}}$

The first integral on the right side has the $g'/g$ form and the second is an $\arctan$ integration.