Integration by partial fraction

We shall describe a method for computing the integral of any rational function, and we shall find that the result can always be expressed in terms of polynomials, rational functions, inverse tangents, and logarithms.

The basic idea of the method is to decompose a given rational function into a sum of simpler fractions (called partial fractions) that can be integrated by the techniques discussed earlier. We shall describe the general procedure by means of a number of simple examples that illustrate all the essential features of the method.

EXAMPLE 1. In this example we begin with two simple fractions, $1/(x-1)$ and $1/(x+3)$ which we know how to integrate, and see what happens when we form a linear combination of these fractions. For example, if we take twice the first fraction plus three times the second, we obtain
MATH

If now, we read this formula from right to left, it tells us that the rational function $r$ given by MATH has been expressed as a linear combination of $1/(x-1)$ and $1/(x+3)$. Therefore, we may evaluate the integral of $r$ by writing
MATH

EXAMPLE 2. The foregoing example suggests a procedure for dealing with integrals of the form MATH For example, to evaluate MATH, we try to express the integral as a linear combination of $1/(x-1)$ and $1/(x+3)$ by writing
MATH
with constants $A$ and $B$ to be determined. If we can choose $A$ and $B$ so that the above equation is an identity, then the integral of the fraction on the left is equal to the sum of the integrals of the simpler fractions on the right. To find $A$ and $B$, we multiply both sides by $(x-1)(x+3)$ to remove the fractions. This gives us
MATH
At this stage there are two methods commonly used to find $A$ and $B$. One method is to equate coefficients of like powers of $x$. This leads to the equations $A+B=2$ and $3A-B=5$. Solving this pair of simultaneous equations, we obtain $A=\frac 74$ and $B=\frac 14.$ The other method involves the substitution of two values of $x$ in the equation and leads to another pair of equations for $A$ and $B$. In this particular case, the presence of the factors $x-1$ and $x+3$ suggests that we use the values $x=1$I and $x=-3$. When we put $x=1$ in the equation, the coefficient of $B$ vanishes, and we find $4A=7$, or $A=\frac 74.$ Similarly, we can make the coefficient of $A$ vanish by putting $x=-3.$ This gives us $-4B=-1$, or $B=1/4$. In any event, we have found values of $A$ and $B$ to satisfy the equation, so we have
MATH

It is clear that the method described in Example 2 also applies to integrals of the form $\int f(x)/g(x)dx$ in which $f$ is a linear polynomial and $g$ is a quadratic polynomial that can be factored into distinct linear factors with real coefficients, say MATH In this case the quotient $\frac{f(x)}{g(x)}$ can be expressed as a linear combination of $1/(x-x_1)$ and $1/(x-x_2)$, and integration of $f(x)/g(x)$ leads to a corresponding combination of the logarithmic terms MATH and MATH.

The foregoing examples involve rational functions $f/g$ in which the degree of the numerator is less than that of the denominator. A rational function with this property is said to be a proper rational function. If $f/g$ is improper, that is, if the degree of $f$ is not less than that of $g$, then we can express $f/g$ as the sum of a polynomial and a proper rational function. In fact, we simply divided by $g$ to obtain
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where $Q$ and $R$ are polynomials (called the quotient and remainder, respectively) such that the remainder has degree less than that of $g$. For example,
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Therefore, in the study of integration technique, there is no loss in generality if we restrict ourselves to proper rational functions, and from now on we consider $\int f(x)/g(x)dx$, where $f$ has degree less than that of $g$.

A general theorem in algebra states that every proper rational function can be expressed as a finite sum of fractions of the forms
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where $k$ and $m$ are positive integers and $A,B,C,a,b,c$ are constants with $b^2-4c<0.$ The condition $b^2-4c<0$ means that the quadratic polynomial $x^2+bx+c$ cannot be factored into linear factors with real coefficients or, what amounts to the same thing, the quadratic equation $x^2+bx+c=0$ has no real roots. Such a quadratic factor is said to be irreducible. When a rational function has been so expressed, we say that it has been decomposed into partial fractions. Therefore the problem of integrating this rational function reduces to that of integrating its partial fractions. These may be easily dealt with by the techniques described in the examples which follow.

We shall not bother to prove that partial-fraction decompositions always exist. Instead, we shall show (by means of examples) how to obtain the partial fractions in specific problems. In each case that arises the partial-fraction decomposition can be verified directly.

It is convenient to separate the discussion into cases depending on the way in which the denominator of the quotient $f(x)/g(x)$ can be factored.

CASE 1. The denominator is a product of distinct linear factors. Suppose that $g(x)$ splits into $n$ distinct linear factors, say
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Now notice that a linear combination of the form
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may be expressed as a single fraction with the common denominator $g(x),$ and the numerator of this fraction will be a polynomial of degree $<n$ involving the $A$'s. Therefore, if we can find $A$'s to make this numerator equal to $f(x)$, we shall have the decomposition
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and the integral of $f(x)/g(x)$ will be equal to MATH In the next example, we work out a case with $n=3.$

EXAMPLE 3. Integrate MATH

Solution. Since MATH, the denominator is a product of distinct linear factors, and we try to find $A_1,A_2,$ and $A_3$ such that
MATH
Clearing the fractions, we obtain
MATH
When $x=0$ we find $-2A_1=-1,$ so $A_1=\frac 12.$ When $x=1,$ we obtain $3A_2=6,A_2=2$, and when $x=-2,$ we find $6A_3=-3,$ or $A_3=\frac 12.$ Therefore we have
MATH

CASE 2. The denominator is a product of linear factors, some of which are repeated. We illustrate this case with an example.

EXAMPLE 4. Integrate MATH.

Solution. Here we try to find $A_1,A_2,A_3$ so that
MATH
We need both $\frac{A_2}{x+1}$ and MATH as well as $\frac{A_1}{x-1}$ in order to get a polynomial of degree two in the numerator and to have as many constants as equations when we try to determine the $A$'s. Clearing the fractions, we obtain
MATH
Substituting $x=1$, we find $4A_1=6$, so $A_1=\frac 32$. When $x=-1,$ we obtain $-2A_3=2$ and $A_3=-1.$ We need one more equation to determine $A_2.$ Since there are no other choices of $x$ that will make any factor vanish, we choose a convenient $x$ that will help to simplify the calculations. For example, the choice $x=0$ leads to the equation $3=A_1-A_2-A_3$ from which we find $A_2=-\frac 12.$ An alternative method is to differentiate both sides of the equation and then substitute a convenient x. Differentiation leads to the equation
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and, if we put $x=1,$we find $0=-2A_2+A_3,$ so MATH as before. Therefore we have found $A$'s to satisfy the equation, so we have
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If, on the left of the equation the factor $(x+1)^3$ had appeared instead of $(x+1)^2$, we would have added an extra term $A_4/(x+1)^3$ on the right. More generally, if a linear factor $x+a$ appears $p$ times in the denominator, then for this factor we must allow for a sum of $p$ terms, namely
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where the $A$'s are constants. A sum of this type is to be used for each repeated linear factor.

CASE 3. The denominator contains irreducible quadratic factors, none of which are repeated.

EXAMPLE 5. Integrate MATH.

Solution. The denominator can be split as the product MATH, where $x^2+x+1$ is irreducible, and we try a decomposition of the form
MATH
In the fraction with denominator $x^2+x+1$, we have used a linear polynomial $Bx+C$ in the numerator in order to have as many constants as equations when we solve for $A,B,C.$ Clearing the fractions and solving for $A,B,$ and $C,$ we find $A=1,B=2,$ and $C=3.$ Therefore we have
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The first integral on the right is MATH. To evaluate the second integral, we write
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If we let $u-=x+1/2$ and MATH, the last integral is
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Therefore, we have
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CASE 4. The denominator contains irreducible quadratic factors, some of which are repeated. Here the situation is analogous to Case 2. In the partial fraction decomposition of $f(x)/g(x)$ we allow first of all, a sum of the form
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for each linear factor, as already described. In addition. if an irreducible quadratic factor $x^2+bx+c$ is repeated $m$ times, we allow a sum of $m$ terms, namely
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where each numerator is linear.

EXAMPLE 6. Integrate MATH

Solution. We write
MATH
Clearing the fractions and solving for $A,B,C,D,$ and $E,$ we find that
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Therefore, we have
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The foregoing examples are typical of what happens in general. The problem of integrating a proper rational function reduces to that of calculating integrals of the forms
MATH
The first integral is MATH if $n=1$ and $(x+a)^{1-n}/(1-n)$ if $n>1.$To treat the other two, we express the quadratic as a sum of two squares by writing
MATH
where $u=x+\frac b2$ and MATH. (This is possible because $4x-b^2>0$.) The substitution $u=x+b/2$ reduces the problem to that of computing
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The first of these is MATH if $m=1$, and MATH if $m>1.$ When $m=1$, the second integral is evaluated by the formula
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The case $m>1$ may be reduced to the case $m=1$ by repeated application of the recursion formula
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which is obtained by integration by parts. This discussion shows that every rational function may he integrated in terms of polynomials, rational functions, inverse tangents, and logarithms.

Integrals which can be transformed into integrals of rational functions

A function of two variables defined by an equation of the form
MATH
is called a polynomial in two variables. The quotient of two such polynomials is called a rational function of two variables. Integrals of the form MATH where $R$ is a rational function of two variables, may be reduced by the substitution $u=\tan \frac 12x$ to integrals of the form $\int r(u)du,$ where $r$ is a rational function of one variable. The latter integral may be evaluated by the techniques just described. We illustrate the method with a particular example.

EXAMPLE 1. Integrate MATH.

Solution. The substitution MATH gives us
MATH

MATH

MATH
and
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Therefore, we have
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where $a=1+\sqrt{2}$ and $b=1-\sqrt{2}$. The method of partial fractions leads to
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and, since $a-b=2\sqrt{2}$, we obtain
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The final answer may be simplified somewhat by using suitable trigonometric identities. First we note that MATH so the numerator of the last fraction is MATH In the denominator we write
MATH
We may combine the term MATH with the arbitrary constant and rewrite as follows:
MATH

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