Integration of rational function with quadratic denominator

Reduction

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!

Indefinite integration for proper fraction with monic denominator

Summary of cases

We assume that we have performed the reductions outlined above. Thus, We consider an integration of the form:

$\int {\frac {(Ax+B)\,dx}{x^{2}+px+q}}$

Note that $A,B,p,q$ are arbitrary constants. They may be zero or nonzero.

Sign of discriminant $p^{2}-4q$	Qualitative description of case	Antiderivative	Functions whose linear combination gives antiderivative	Description of new constants in terms of $p,q$
positive	denominator has distinct roots $\alpha _{1},\alpha _{2}$ factors as $(x-\alpha _{1})(x-\alpha _{2})$	${\frac {A\alpha _{1}+B}{\alpha _{1}-\alpha _{2}}}\ln |x-\alpha _{1}|+{\frac {A\alpha _{2}+B}{\alpha _{2}-\alpha _{1}}}\ln |x-\alpha _{2}|+C$	$\ln |x-\alpha _{1}|$ $\ln |x-\alpha _{2}|$	$\!\alpha _{1}={\frac {-p-{\sqrt {p^{2}-4q}}}{2}},\alpha _{2}={\frac {-p+{\sqrt {p^{2}-4q}}}{2}}$
zero	denominator has repeated root $\sigma$ factors as $(x-\sigma )^{2}$	$A\ln |x-\sigma |-{\frac {A\sigma +B}{x-\sigma }}+C$	$\!\ln |x-\sigma |$ ${\frac {1}{x-\sigma }}$	$\!\sigma =-p/2$
negative	denominator is of the form $(x-\beta )^{2}+\gamma ^{2}$	${\frac {A}{2}}\ln((x-\beta )^{2}+\gamma ^{2})+{\frac {A\beta +B}{\gamma }}\arctan \left({\frac {x-\beta }{\gamma }}\right)+C$	$\ln((x-\beta )^{2}+\gamma ^{2})$ $\arctan \left({\frac {x-\beta }{\gamma }}\right)$	$\!\beta ={\frac {-p}{2}},\gamma ={\sqrt {q-(p^{2}/4)}}$

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:

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

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

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.

Definite integration

We discuss here only the reduced case that we deal with for the indefinite integration.

Case that the denominator has distinct linear factors

Using the same notation as for the indefinite integration, we have an integration of the form:

$\!{\frac {(Ax+B)\,dx}{(x-\alpha _{1})(x-\alpha _{2})}}$

The antiderivative is:

${\frac {A\alpha _{1}+B}{\alpha _{1}-\alpha _{2}}}\ln |x-\alpha _{1}|+{\frac {A\alpha _{2}+B}{\alpha _{2}-\alpha _{1}}}\ln |x-\alpha _{2}|+C$

where, if the quadratic in the denominator originally is $x^{2}+px+q$:

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

The integrand is not defined at the points $\alpha _{1},\alpha _{2}$. In fact, the domain of the function is a union of three separate open intervals: $(-\infty ,\alpha _{1})$, $(\alpha _{1},\alpha _{2})$, and $(\alpha _{2},\infty )$.

A generic antiderivative for the function across all the intervals may have different values for the constant $C$ on each interval.

Moreover, we can drop the absolute value signs and be specific about the inputs to the $\ln$s within each interval:

Interval	Antiderivative on the interval
$(-\infty ,\alpha _{1})$	${\frac {A\alpha _{1}+B}{\alpha _{1}-\alpha _{2}}}\ln(\alpha _{1}-x)+{\frac {A\alpha _{2}+B}{\alpha _{2}-\alpha _{1}}}\ln(\alpha _{2}-x)+C$
$(\alpha _{1},\alpha _{2})$	${\frac {A\alpha _{1}+B}{\alpha _{1}-\alpha _{2}}}\ln(x-\alpha _{1})+{\frac {A\alpha _{2}+B}{\alpha _{2}-\alpha _{1}}}\ln(\alpha _{2}-x)+C$
$(\alpha _{2},\infty )$	${\frac {A\alpha _{1}+B}{\alpha _{1}-\alpha _{2}}}\ln(x-\alpha _{1})+{\frac {A\alpha _{2}+B}{\alpha _{2}-\alpha _{1}}}\ln(x-\alpha _{2})+C$

The integral can be computed to give a finite numerical value on any interval properly contained completely within one of these intervals. We now address the question of whether we can compute improper integrals, i.e., integrals where one of the limits is one of the values $-\infty ,\alpha _{1},\alpha _{2},\infty$.

The answer is as follows:

• An improper integral extending to $\alpha _{1}$ will be finite only if $A\alpha _{1}+B=0$. Note that if that were the case, the rational function would not have been in a reduced form, i.e., there would have been a common factor between the numerator and denominator. In fact, in this case, the original rational function would have a removable discontinuity at $\alpha _{1}$.
• An improper integral extending to $\alpha _{2}$ will be finite only if $A\alpha _{2}+B=0$. Note that if that were the case, the rational function would not have been in a reduced form, i.e., there would have been a common factor between the numerator and denominator. In fact, in this case, the original rational function would have a removable discontinuity at $\alpha _{1}$.
• An improper integral extending to $-\infty$ will be finite only if $A=0$, so the numerator is constant. This agrees with the degree difference test. For the antiderivative given above, the limiting value at $-\infty$ is 0.
• An improper integral extending to $+\infty$ will be finite only if $A=0$, so the numerator is constant. This agrees with the degree difference test. For the antiderivative given above, the limiting value at $+\infty$ is 0.