[CALCULUS] Rational functions

This page integrates rational functions, which are ratios of polynomials. If the fraction is improper, divide first; the polynomial part integrates immediately and the proper remainder decomposes uniquely into partial fractions. Linear factors lead to logarithms and negative powers, while non-repeated irreducible quadratics lead to logarithms and inverse tangents. Repeated quadratics may also need a reduction step or a trigonometric substitution.

Other integrands belong elsewhere: trigonometric powers in trigonometric integrals, quadratic surds in trigonometric and hyperbolic substitutions, and other algebraic surds in rationalising substitutions.

Classifying and dividing rational functions

[DEFINITION]

Rational function

A function $f$ is a rational function if it is of the form $f(x)=\frac{P(x)}{Q(x)}$ , where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial. Its real domain excludes the zeros of $Q$ .

[DEFINITION]

Proper rational function

A rational function $f$ is proper if $\deg P<\deg Q$ . If $f$ is not proper, long division writes it as a polynomial plus a proper rational function,

\frac{P(x)}{Q(x)}=p(x)+\frac{R(x)}{Q(x)},

with $\frac{R(x)}{Q(x)}$ proper.

The polynomial $p$ integrates termwise. Only the proper remainder needs a further decomposition.

[EXAMPLE]

Which of these are rational?

$\frac{2x+1}{3x^2-4}$ is rational and proper: $\deg(2x+1)=1<2=\deg(3x^2-4)$ .
$\frac{\cos x}{x^2+1}$ is not rational: $\cos x$ is not a polynomial.
$\frac{x^2}{x-1}$ is rational and improper: $\deg(x^2)=2>1=\deg(x-1)$ .
$\frac{\sqrt{x}}{x^2+3}$ is not rational: $\sqrt{x}$ is not a polynomial.

[routine]

[1]:Rational function

[EXAMPLE]

Polynomial plus proper remainder

Write each rational function as a polynomial plus a proper rational function.

For $\frac{4x-5}{2x+3}$ , the leading-term ratio is $2$ and $2(2x+3)=4x+6$ , so

(4x-5)-(4x+6)=-11.

Hence

\frac{4x-5}{2x+3}=2-\frac{11}{2x+3}.

For $\frac{x^3-x+1}{x^2+3x-1}$ , divide. The first quotient term is $x$ and

x(x^2+3x-1)=x^3+3x^2-x,

(x^3-x+1)-(x^3+3x^2-x)=-3x^2+1.

The next quotient term is $-3$ and

-3(x^2+3x-1)=-3x^2-9x+3,

(-3x^2+1)-(-3x^2-9x+3)=9x-2.

Hence

\frac{x^3-x+1}{x^2+3x-1}=x-3+\frac{9x-2}{x^2+3x-1}.

[exam]

[1]:Proper rational function

Decomposing into partial fractions

Once the remainder is proper, cancel any common factors and factor its denominator over $\mathbb{R}$ . Cancellation simplifies the formula only on the original domain: any zero of the original denominator remains an excluded point when choosing an integration interval. A separate partial-fraction term is needed for every power of every factor; omitting a power can make the coefficient equations impossible to solve.

[THEOREM]

Partial fractions decomposition

For a fixed denominator after cancellation and factorisation, every proper rational function $f$ has a unique partial-fraction expansion with terms of the form

\frac{A}{(x-a)^k},\qquad \frac{Bx+C}{(x^2+bx+c)^k},

where the quadratic $x^2+bx+c$ is irreducible over the real numbers (i.e. $b^2-4c<0$ ).

The integers $k$ run up to the multiplicity of each factor. Uniqueness means the linear equations for the undetermined coefficients have exactly one solution. The real irreducible quadratics arise by pairing a non-real complex root with its conjugate, as in real linear and quadratic factorisation.

For $q(x)=ax^2+bx+c$ , the quantity $\Delta=b^2-4ac$ is its discriminant. If $\Delta<0$ , the quadratic has no real roots and is irreducible over $\mathbb R$ ; this is the quadratic case in the partial-fraction forms above.

[METHOD]

Partial fraction terms

Given a proper rational function, each irreducible factor of the denominator contributes terms as follows.

Linear factor $ax+b$ : $\frac{A}{ax+b}$ .
Repeated linear factor $(ax+b)^k$ : $\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_k}{(ax+b)^k}$ .
Irreducible quadratic factor $ax^2+bx+c$ with $a\ne0$ : $\frac{Ax+B}{ax^2+bx+c}$ .
Repeated irreducible quadratic factor $(ax^2+bx+c)^k$ : $\frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\cdots+\frac{A_kx+B_k}{(ax^2+bx+c)^k}$ .

Here each factor has a general leading coefficient $a$ , whereas the decomposition theorem used the monic quadratic $x^2+bx+c$ , irreducible when $b^2-4c<0$ .

[METHOD]

Integrate a rational function

To integrate a rational function on an interval containing no zero of its denominator:

If the function is improper, use polynomial division to write it as a polynomial plus a proper rational function and integrate the polynomial.
Decompose the proper part into partial fractions.
Integrate each linear-factor term using $u=ax+b$ : its first power gives $\frac{A}{a}\ln|ax+b|$ , while a power $k>1$ gives $\frac{A(ax+b)^{1-k}}{a(1-k)}$ .
For $q(x)=ax^2+bx+c$ and a term $\int\frac{Bx+C}{q(x)^k}\,dx$ , split the numerator into a multiple of $q'(x)=2ax+b$ and a constant remainder:

\frac{Bx+C}{q(x)^k} =\frac{B}{2a}\frac{q'(x)}{q(x)^k}+\frac{D}{q(x)^k}, \qquad D=C-\frac{Bb}{2a}.

Then integrate the derivative term directly and complete the square in the constant remainder.

The derivative term is immediate:

\int\frac{g'(x)}{g(x)}\, dx=\ln|g(x)|+K,

and, for $k>1$ ,

\int\frac{g'(x)}{g(x)^k}\,dx =\frac{g(x)^{1-k}}{1-k}+K.

For the constant term, complete the square. When $k=1$ , this produces the standard inverse-tangent integral

\int\frac{1}{h^2+x^2}\, dx=\frac{1}{h}\tan^{-1}\left(\frac{x}{h}\right)+K, \qquad h>0.

For $k>1$ , reduce the remaining power or use a trigonometric substitution, as in the repeated-quadratic example below.

Integrating linear factors

[EXAMPLE]

Distinct linear factors

Find $\int\frac{2x^2+2x-6}{(x-1)(x+1)(x-2)}\, dx$ .

The denominator is a product of distinct linear factors, so

\frac{2x^2+2x-6}{(x-1)(x+1)(x-2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}.

Call this identity $(*)$ . Heaviside's cover-up method finds $A$ , $B$ and $C$ . Multiplying $(*)$ by $x-1$ yields

\frac{2x^2+2x-6}{(x+1)(x-2)}=A+\frac{B(x-1)}{x+1}+\frac{C(x-1)}{x-2}.

Identity $(*)$ holds for $x\neq 1,-1,2$ , but after multiplying through by $x-1$ both sides are defined at $x=1$ , so setting $x=1$ is legitimate:

\frac{2(1)^2+2(1)-6}{(1+1)(1-2)}=\frac{-2}{2\cdot(-1)}=1,

hence $A=1$ .

Similarly, multiplying $(*)$ by $x+1$ and setting $x=-1$ gives

B=\left.\frac{2x^2+2x-6}{(x-1)(x-2)}\right|_{x=-1}=-1,

and doing the same with $x-2$ and $x=2$ gives

C=\left.\frac{2x^2+2x-6}{(x-1)(x+1)}\right|_{x=2}=2.

Hence

\int\frac{2x^2+2x-6}{(x-1)(x+1)(x-2)}\, dx=\int\frac{dx}{x-1}-\int\frac{dx}{x+1}+\int\frac{2}{x-2}\, dx

=\ln|x-1|-\ln|x+1|+2\ln|x-2|+C.

[exam]

[1]:Partial fractions decomposition

[EXAMPLE]

Repeated linear factors

Find $I=\int\frac{8x+9}{(x-2)(x+3)^2}\, dx$ .

The factor $x+3$ is repeated, so

\frac{8x+9}{(x-2)(x+3)^2}=\frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^2}

(or, equivalently, $\frac{A}{x-2}+\frac{Bx+C}{(x+3)^2}$ ). With the three-term form, the cover-up method from the distinct linear factors example gives $A$ by multiplying by $x-2$ and setting $x=2$ :

A=\left.\frac{8x+9}{(x+3)^2}\right|_{x=2}=\frac{16+9}{25}=1.

Multiplying by $(x+3)^2$ and setting $x=-3$ gives

C=\left.\frac{8x+9}{x-2}\right|_{x=-3}=\frac{-24+9}{-5}=3.

Cover-up does not give $B$ : multiplying by $x+3$ leaves a term $\frac{C}{x+3}$ , so $x=-3$ cannot be substituted. Instead, insert the known $A$ and $C$ ,

\frac{8x+9}{(x-2)(x+3)^2}=\frac{1}{x-2}+\frac{B}{x+3}+\frac{3}{(x+3)^2}

and set $x=0$ :

\frac{9}{-18}=-\frac{1}{2}+\frac{B}{3}+\frac{1}{3}\implies B=-1.

Hence

I=\int\frac{dx}{x-2}-\int\frac{dx}{x+3}+\int\frac{3}{(x+3)^2}\, dx=\ln|x-2|-\ln|x+3|-\frac{3}{x+3}+C.

This formula is valid on any interval that contains neither pole, $x=2$ nor $x=-3$ ; the constant need not be the same on different such intervals.

[exam]

[1]:Partial fractions decomposition

Integrating irreducible quadratics

For an irreducible quadratic denominator, rewrite the numerator around the derivative of the quadratic. This splits off a logarithm and the constant remainder becomes an inverse tangent after completing the square.

[EXAMPLE]

Linear numerator over a quadratic

Find $I=\int\frac{2x+1}{x^2+4x+5}\, dx$ .

The quadratic $x^2+4x+5$ is irreducible ( $4^2-4\cdot5=-4<0$ ), so write the numerator in terms of the derivative $2x+4$ of the denominator:

2x+1=(2x+4)-3.

Thus

I=\int\frac{2x+4}{x^2+4x+5}\, dx-\int\frac{3}{x^2+4x+5}\, dx.

The first integrand is $g'/g$ . Completing the square in the second, $x^2+4x+5=(x+2)^2+1$ , gives

I=\ln|x^2+4x+5|-\int\frac{3}{(x+2)^2+1}\, dx=\ln|x^2+4x+5|-3\tan^{-1}(x+2)+C.

[exam]

[1]:Integrate a rational function

[EXAMPLE]

A repeated irreducible quadratic

Find $I=\int\frac{2x^2-x}{(x^2+1)^2}\, dx$ .

Decompose as

\frac{2x^2-x}{(x^2+1)^2}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}.

Multiplying by $(x^2+1)^2$ ,

2x^2-x=(Ax+B)(x^2+1)+Cx+D=Ax^3+Bx^2+(A+C)x+(B+D).

Comparing coefficients, $A=0$ , $B=2$ , $A+C=-1$ so $C=-1$ and $B+D=0$ so $D=-2$ . Hence

I=\int\frac{2}{x^2+1}\, dx-\int\frac{x+2}{(x^2+1)^2}\, dx=2\tan^{-1}(x)-\frac{1}{2}\int\frac{2x}{(x^2+1)^2}\, dx-2\int\frac{1}{(x^2+1)^2}\, dx.

The middle integral is $\int g'/g^2$ with $g(x)=x^2+1$ , so

I=2\tan^{-1}(x)+\frac{1}{2}(x^2+1)^{-1}-2\int\frac{1}{(x^2+1)^2}\, dx.

For the remaining integral, let $x=\tan\theta$ , so $dx=\sec^2\theta\, d\theta$ (the $x=a\tan\theta$ row of the trigonometric and hyperbolic substitutions, with $a=1$ ). Then $\tan^2\theta+1=\sec^2\theta$ and

\int\frac{1}{(x^2+1)^2}\, dx=\int\frac{\sec^2\theta}{(\sec^2\theta)^2}\, d\theta=\int\cos^2\theta\, d\theta.

Using $\cos^2\theta=\frac{1+\cos 2\theta}{2}$ , as in trigonometric integrals,

\int\cos^2\theta\, d\theta=\frac{\theta}{2}+\frac{\sin(2\theta)}{4}+C=\frac{1}{2}\tan^{-1}(x)+\frac{1}{4}\sin(2\tan^{-1}(x))+C.

Altogether,

I=\tan^{-1}(x)+\frac{1}{2}(x^2+1)^{-1}-\frac{1}{2}\sin(2\tan^{-1}(x))+C.

To rewrite the sine, let $\alpha=\tan^{-1}(x)$ , so $\tan\alpha=x$ and $\cos\alpha>0$ . From $1+\tan^2\alpha=\sec^2\alpha$ and the sign of $\cos\alpha$ ,

\sin\alpha=\frac{x}{\sqrt{x^2+1}},\qquad \cos\alpha=\frac{1}{\sqrt{x^2+1}}.

The double-angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$ then gives

\sin(2\tan^{-1}(x))=2\left(\frac{x}{\sqrt{x^2+1}}\right)\left(\frac{1}{\sqrt{x^2+1}}\right)=\frac{2x}{x^2+1}.

Thus

I=\tan^{-1}(x)+\frac{1-2x}{2(x^2+1)}+C.

[exam]

Rational functions

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