Partial Fraction Decomposition

Proper and Improper Rational Expressions

A rational expression is any quotient of form $ $ , where f and g are polynomial expressions. (Here it is assumed that f and g have real coefficients.) If the degree of f is less than the degree of g, the rational expression is called proper, otherwise improper. An improper rational expression can always be written, using the long division scheme (Chapter 14), as a polynomial plus a proper rational expression.

Partial Fraction Decomposition

Any polynomial $ g(x) $ can, theoretically, be written as the product of one or more linear and quadratic factors, where the quadratic factors have no real zeros (irreducible quadratic factors). It follows that any proper rational expression with denominator $ g(x) $ can be written as a sum of one or more proper rational expressions, each having a denominator that is a power of a polynomial with degree less than or equal to 2. This sum is called the partial fraction decomposition of the rational expression.

EXAMPLE 32.1 $ $ is an improper rational expression. It can be rewritten as the sum of a polynomial and a proper rational expression: $ = x - 1 + $ .

EXAMPLE 32.2 $ $ is a proper rational expression. Since its denominator factors as $ x^{2}+x=x(x+1) $ , the partial fraction decomposition of $ $ is $ =+ $ , as can be verified by addition:

\[ \frac{1}{x}+\frac{1}{x+1}=\frac{x+1}{x(x+1)}+\frac{x}{x(x+1)}=\frac{2x+1}{x^{2}+x} \]

EXAMPLE 32.3 $ $ is already in partial fraction decomposed form, since the denominator is quadratic and has no real zeros.

Procedure for Finding the Partial Fraction Decomposition

Procedure for finding the partial fraction decomposition of a rational expression:

1. If the expression is proper, go to step 2. If the expression is improper, divide to obtain a polynomial plus a proper rational expression and apply the following steps to the proper expression $ f(x)/g(x) $ .

2. Write the denominator as a product of powers of linear factors of form $ (ax + b)^{m} $ and irreducible quadratic factors of form $ (ax^{2} + bx + c)^{n} $ .

3. For each factor $ (ax + b)^{m} $ , write a partial fraction sum of form:

\[ \frac{A_{1}}{ax+b}+\frac{A_{2}}{(ax+b)^{2}}+\ \cdots\ +\frac{A_{m}}{(ax+b)^{m}} \]

where the $ A_{i} $ are as yet to be determined unknown coefficients.

4. For each factor $ (ax^{2} + bx + c)^{n} $ , write a partial fraction sum of form:

\[ \frac{B_{1}x+C_{1}}{ax^{2}+bx+c}+\frac{B_{2}x+C_{2}}{(ax^{2}+bx+c)^{2}}+\ \cdots\ +\frac{B_{n}x+C_{n}}{(ax^{2}+bx+c)^{n}} \]

where the \(B_{i}\) and \(C_{i}\) are as yet to be determined unknown coefficients.

5. Set $ f(x)/g(x) $ equal to the sum of the partial fractions from steps 4 and 5. Eliminate the denominator $ g(x) $ by multiplying both sides to obtain the basic equation for the unknown coefficients.

6. Solve the basic equation for the unknown coefficients $ A_{i}, B_{j} $ , and $ C_{j} $ .

General Method for Solving the Basic Equation

1. Expand both sides.

2. Collect terms in each power of x.

3. Equate coefficients of each power of x.

4. Solve the linear system in the unknowns $ A_{i} $ , $ B_{i} $ , and $ C_{i} $ that results.

EXAMPLE 32.4 Find the partial fraction decomposition of $ $ .

This is a proper rational expression. The denominator $ x^{2}-1 $ factors as $ (x-1)(x+1) $ . Therefore, there are only two partial fraction sums, one with denominator x-1 and the other with denominator $ x+1 $ . Then set

\[ \frac{4}{x^{2}-1}=\frac{A_{1}}{x-1}+\frac{A_{2}}{x+1} \]

Multiply both sides by $ x^{2}-1 $ to obtain the basic equation

\[ 4=A_{1}(x+1)+A_{2}(x-1) \]

Expanding yields

\[ 4=A_{1}x+A_{1}+A_{2}x-A_{2} \]

Collecting terms in each power of x yields

\[ 0x+4=(A_{1}+A_{2})x+(A_{1}-A_{2}) \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation must be equal; hence:

\[ \begin{aligned}&A_{1}+A_{2}=0\\&A_{1}-A_{2}=4\\ \end{aligned} \]

The system has one solution: \(A_{1}=2\), \(A_{2}=-2\). Hence the partial fraction decomposition is

\[ \frac{4}{x^{2}-1}=\frac{2}{x-1}+\frac{-2}{x+1} \]

Alternate Method

Alternate method for solving the basic equation: Instead of expanding both sides of the basic equation, substitute values for x into the equation. If, and only if, all partial fractions have distinct linear denominators, if the values chosen are the distinct zeros of these expressions, the values of the $ A_{i} $ will be found immediately. In other situations, there will not be enough of these zeros to determine all the unknowns. Other values of x may be chosen and the resulting system of equations solved, but in these situations the alternative method is not preferred.

EXAMPLE 32.5 Use the alternative method for the previous example.

The basic equation is

\[ 4=A_{1}(x+1)+A_{2}(x-1) \]

Substitute x = 1, then it follows that:

\[ 4=A_{1}(1+1)+A_{2}(1-1) \]

\[ 4=2A_{1} \]

\[ A_{1}=2 \]

Now substitute x = -1, then it follows that:

\[ 4=A_{1}(-1+1)+A_{2}(-1-1) \]

\[ 4=-2A_{2} \]

\[ A_{2}=-2 \]

This yields the same result as before.

SOLVED PROBLEMS

32.1. Find the partial fraction decomposition of $ $ .

This is a proper rational expression. The denominator factors as follows:

\[ x^{3}-x=x(x^{2}-1)=x(x-1)(x+1) \]

Thus there are three partial fraction sums, one each with denominator x, x - 1, and $ x + 1 $ . Set

\[ \frac{x^{2}+7x-2}{x^{3}-x}=\frac{A_{1}}{x}+\frac{A_{2}}{x-1}+\frac{A_{3}}{x+1} \]

Multiplying both sides by $ x^{3}-x=x(x-1)(x+1) $ yields

\[ \begin{aligned}(x^{3}-x)\frac{x^{2}+7x-2}{x^{3}-x}&=x(x-1)(x+1)\frac{A_{1}}{x}+x(x-1)(x+1)\frac{A_{2}}{x-1}+x(x-1)(x+1)\frac{A_{3}}{x+1}\\x^{2}+7x-2&=A_{1}(x-1)(x+1)+A_{2}x(x+1)+A_{3}x(x-1)\end{aligned} \]

This is the basic equation. Since all the partial fractions have linear denominators, it is more efficient to apply the alternate method. Substitute for x, in turn, the zeros of the denominator $ x(x-1)(x+1) $ .

\[ x=0; \]

\[ \begin{aligned}-2&=A_{1}(-1)(1)+A_{2}(0)(1)+A_{3}(0)(-1)\\-2&=-A_{1}\\A_{1}&=2\end{aligned} \]

x = 1:

\[ \begin{aligned}1^{2}+7\cdot1-2&=A_{1}(1-1)(1+1)+A_{2}(1)(1+1)+A_{3}(1)(1-1)\\6&=2A_{2}\\A_{2}&=3\end{aligned} \]

\[ x = -1: \]

\[ \begin{aligned}(-1)^{2}+7(-1)-2&=A_{1}(-1-1)(-1+1)+A_{2}(-1)(-1+1)+A_{3}(-1)(-1-1)\\-8&=2A_{3}\\A_{3}&=-4\end{aligned} \]

Hence the partial fraction decomposition is

\[ \frac{x^{2}+7x-2}{x^{3}-x}=\frac{2}{x}+\frac{3}{x-1}+\frac{-4}{x+1} \]

32.2. Find the partial fraction decomposition of $ $ .

This is an improper expression. Use the long division scheme to rewrite it as:

\[ x+1+\frac{5x-8}{6x^{2}-x-2} \]

The denominator factors as $ (3x - 2)(2x + 1) $ . Thus there are two partial fraction sums, one with denominator 3x - 2 and the other with denominator $ 2x + 1 $ . Set

\[ \frac{5x-8}{6x^{2}-x-2}=\frac{A_{1}}{3x-2}+\frac{A_{2}}{2x+1} \]

Multiply both sides by $ 6x^{2}-x-2=(3x-2)(2x+1) $ to obtain

\[ 5x-8=A_{1}(2x+1)+A_{2}(3x-2) \]

This is the basic equation. Since the zeros of the denominator involve fractions, the alternate method does not seem attractive. Expanding yields

\[ 5x-8=2A_{1}x+A_{1}+3A_{2}x-2A_{2} \]

Collecting terms in each power of x yields

\[ 5x-8=(2A_{1}+3A_{2})x+(A_{1}-2A_{2}) \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation must be equal; hence:

\[ \begin{aligned}2A_{1}+3A_{2}&=5\\A_{1}-2A_{2}&=-8\end{aligned} \]

The only solution of this system is $ A_{1} = -2 $ , $ A_{2} = 3 $ . Hence the partial fraction decomposition is

\[ \frac{6x^{3}+5x^{2}+2x-10}{6x^{2}-x-2}=x+1+\frac{-2}{3x-2}+\frac{3}{2x+1} \]

32.3. Find the partial fraction decomposition of $ $ .

This is an improper expression. Use the long division scheme to rewrite it as:

\[ -x+\frac{3x^{3}+5x^{2}+6x+6}{x^{4}+x^{3}} \]

The denominator factors as $ x^{3}(x + 1) $ . The first factor is referred to as a repeated linear factor; one partial fraction sum must be considered for each power of x from 1 to 3. Set

\[ \frac{3x^{3}+5x^{2}+6x+6}{x^{4}+x^{3}}=\frac{A_{1}}{x}+\frac{A_{2}}{x^{2}}+\frac{A_{3}}{x^{3}}+\frac{A_{4}}{x+1} \]

Multiply both sides by $ x^{4} + x^{3} = x^{3}(x + 1) $ to obtain

\[ 3x^{3}+5x^{2}+6x+6=A_{1}x^{2}(x+1)+A_{2}x(x+1)+A_{3}(x+1)+A_{4}x^{3} \]

This is the basic equation. Expanding yields

\[ 3x^{3}+5x^{2}+6x+6=A_{1}x^{3}+A_{1}x^{2}+A_{2}x^{2}+A_{2}x+A_{3}x+A_{3}+A_{4}x^{3} \]

Collecting terms in each power of x yields

\[ 3x^{3}+5x^{2}+6x+6=(A_{1}+A_{4})x^{3}+(A_{1}+A_{2})x^{2}+(A_{2}+A_{3})x+A_{3} \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation must be equal, hence:

\[ \begin{aligned}A_{1}+A_{4}&=3\\A_{1}+A_{2}&=5\\A_{2}+A_{3}&=6\\A_{3}&=6\end{aligned} \]

The only solution of this system is $ A_{1}=5 $ , $ A_{2}=0 $ , $ A_{3}=6 $ , $ A_{4}=-2 $ . Hence the partial fraction decomposition is

\[ \frac{-x^{5}-x^{4}+3x^{3}+5x^{2}+6x+6}{x^{4}+x^{3}}=-x+\frac{5}{x}+\frac{6}{x^{3}}+\frac{-2}{x+1} \]

32.4. Find the partial fraction decomposition of $ $ .

This is a proper rational expression. The denominator factors as follows:

\[ x^{4}-1=(x^{2}-1)(x^{2}+1)=(x-1)(x+1)(x^{2}+1) \]

Thus there are three partial fraction sums, one each with denominator $ x - 1 $ , $ x + 1 $ , and $ x^{2} + 1 $ . Note that the irreducible quadratic denominator $ x^{2} + 1 $ requires a numerator of the form $ B_{1}x + C_{1} $ , that is, a linear rather than a constant expression. Set

\[ \frac{x^{3}-x^{2}+9x-1}{x^{4}-1}=\frac{A_{1}}{x-1}+\frac{A_{2}}{x+1}+\frac{B_{1}x+C_{1}}{x^{2}+1} \]

Multiply both sides by $ x^{{4}-1=(x-1)(x+1)(x}{2}+1) $ to obtain

\[ x^{3}-x^{2}+9x-1=A_{1}(x+1)(x^{2}+1)+A_{2}(x-1)(x^{2}+1)+(B_{1}x+C_{1})(x-1)(x+1) \]

This is the basic equation. Expanding yields

\[ x^{3}-x^{2}+9x-1=A_{1}x^{3}+A_{1}x^{2}+A_{1}x+A_{1}+A_{2}x^{3}-A_{2}x^{2}+A_{2}x-A_{2}+B_{1}x^{3}+C_{1}x^{2}-B_{1}x-C_{1} \]

Collecting terms in each power of x yields

\[ x^{3}-x^{2}+9x-1=(A_{1}+A_{2}+B_{1})x^{3}+(A_{1}-A_{2}+C_{1})x^{2}+(A_{1}+A_{2}-B_{1})x+A_{1}-A_{2}-C_{1} \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation must be equal, hence:

\[ \begin{aligned}&A_{1}+A_{2}+B_{1}=1\\&A_{1}-A_{2}+C_{1}=-1\\&A_{1}+A_{2}-B_{1}=9\\&A_{1}-A_{2}-C_{1}=-1\end{aligned} \]

The only solution of this system is $ A_{1}=2 $ , $ A_{2}=3 $ , $ B_{1}=-4 $ , $ C_{1}=0 $ . Hence the partial fraction decomposition is

\[ \frac{x^{3}-x^{2}+9x-1}{x^{4}-1}=\frac{2}{x-1}+\frac{3}{x+1}+\frac{-4x}{x^{2}+1} \]

32.5. Find the partial fraction decomposition of $ $

This is a proper rational expression. The denominator factors as $ x^{4} + 10x^{2} + 9 = (x^{2} + 1)(x^{2} + 9) $ . There are only two partial fractions, one each with denominator $ x^{2} + 1 $ and $ x^{2} + 9 $ . Each irreducible quadratic denominator requires a linear, not a constant numerator. Set

\[ \frac{5x^{3}-4x^{2}+21x-28}{x^{4}+10x^{2}+9}=\frac{B_{1}x+C_{1}}{x^{2}+1}+\frac{B_{2}x+C_{2}}{x^{2}+9} \]

Multiply both sides by $ x^{4} + 10x^{2} + 9 = (x^{2} + 1)(x^{2} + 9) $ to obtain

\[ 5x^{3}-4x^{2}+21x-28=(B_{1}x+C_{1})(x^{2}+9)+(B_{2}x+C_{2})(x^{2}+1) \]

This is the basic equation. Expanding yields

\[ 5x^{3}-4x^{2}+21x-28=B_{1}x^{3}+C_{1}x^{2}+9B_{1}x+9C_{1}+B_{2}x^{3}+C_{2}x^{2}+B_{2}x+C_{2} \]

Collecting terms in each power of x yields

\[ 5x^{3}-4x^{2}+21x-28=(B_{1}+B_{2})x^{3}+(C_{1}+C_{2})x^{2}+(9B_{1}+B_{2})x+9C_{1}+C_{2} \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation must be equal, hence:

\[ \begin{aligned}B_{1}+B_{2}&=5\\C_{1}+C_{2}&=-4\\9B_{1}+B_{2}&=21\\9C_{1}+C_{2}&=-28\end{aligned} \]

The only solution of this system is $ B_{1}=2 $ , $ B_{2}=3 $ , $ C_{1}=-3 $ , $ C_{2}=-1 $ . Hence the partial fraction decomposition is

\[ \frac{5x^{3}-4x^{2}+21x-28}{x^{4}+10x^{2}+9}=\frac{2x-3}{x^{2}+1}+\frac{3x-1}{x^{2}+9} \]

32.6. A common error in setting up a partial fraction sum is to assign a constant numerator to a partial fraction with an irreducible quadratic denominator. Explain what would happen in the previous problem as a result of this error.

Assume the incorrect partial fraction sum

\[ \frac{5x^{3}-4x^{2}+21x-28}{x^{4}+10x^{2}+9}=\frac{A_{1}}{x^{2}+1}+\frac{A_{2}}{x^{2}+9} \]

is set up. Multiplying both sides by $ x^{4} + 10x^{2} + 9 = (x^{2} + 1)(x^{2} + 9) $ would yield

\[ 5x^{3}-4x^{2}+21x-28=A_{1}(x^{2}+9)+A_{2}(x^{2}+1) \]

Expanding this incorrect basic equation would yield

\[ 5x^{3}-4x^{2}+21x-28=A_{1}x^{2}+9A_{1}+A_{2}x^{2}+A_{2} \]

Collecting terms in each power of x would yield

\[ 5x^{3}-4x^{2}+21x-28=(A_{1}+A_{2})x^{2}+9A_{1}+A_{2} \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation would have to be equal, but this is impossible; for example, the coefficient of $ x^{3} $ on the left is 5, but on the right it is 0. So the problem has been tackled incorrectly.

32.7. Find the partial fraction decomposition of $ $

This is a proper rational expression. The denominator factors as $ x^{4} + 8x^{2} + 16 = (x^{2} + 4)^{2} $ . This is referred to as a repeated quadratic factor; one partial fraction sum must be considered for both $ x^{2} + 4 $ and $ (x^{2} + 4)^{2} $ . Each irreducible quadratic denominator requires a linear, not a constant, numerator. Set

\[ \frac{3x^{3}+14x-3}{x^{4}+8x^{2}+16}=\frac{B_{1}x+C_{1}}{x^{2}+4}+\frac{B_{2}x+C_{2}}{(x^{2}+4)^{2}} \]

Multiply both sides by $ x^{4} + 8x^{2} + 16 = (x^{2} + 4)^{2} $ to obtain

\[ 3x^{3}+14x-3=(B_{1}x+C_{1})(x^{2}+4)+B_{2}x+C_{2} \]

This is the basic equation. Expanding yields

\[ 3x^{3}+14x-3=B_{1}x^{3}+C_{1}x^{2}+4B_{1}x+4C_{1}+B_{2}x+C_{2} \]

Collecting terms in each power of x yields

\[ 3x^{3}+14x-3=B_{1}x^{3}+C_{1}x^{2}+(4B_{1}+B_{2})x+4C_{1}+C_{2} \]

For this to hold for all x, the coefficients of each power of x on both sides of the equation must be equal, hence:

\[ B_{1}=3 \]

\[ C_{1}=0 \]

\[ 4B_{1}+B_{2}=14 \]

\[ 4C_{1}+C_{2}=-3 \]

The only solution of this system is $ B_{1}=3 $ , $ B_{2}=2 $ , $ C_{1}=0 $ , $ C_{2}=-3 $ . Hence the partial fraction decomposition is

\[ \frac{3x^{3}+14x-3}{x^{4}+8x^{2}+16}=\frac{3x}{x^{2}+4}+\frac{2x-3}{(x^{2}+4)^{2}} \]

SUPPLEMENTARY PROBLEMS

32.8. Find the partial fraction decomposition of $ $ .

Ans. $ + $

32.9. Find the partial fraction decomposition of $ $ .

Ans. $ + $

32.10. Find the partial fraction decomposition of $ $ .

Ans. $ x^{2} + 3x - 1 + + $

32.11. Find the partial fraction decomposition of $ $ .

Ans. $ ++ $

32.12. Find the partial fraction decomposition of $ $ .

Ans. $ + + $

32.13. Find the partial fraction decomposition of $ $ .

Ans. $ ++ $

32.14. Find the partial fraction decomposition of $ $ .

Ans. $ x^{{2}++\frac{3}{x}{3}}+ $

32.15. Find the partial fraction decomposition of $ $ .

Ans. $ x + + $

32.16. Find the partial fraction decomposition of $ $ .

\[ Ans.\quad\frac{2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{-3}{x^{2}+1} \]

32.17. Find the partial fraction decomposition of $ $ .

\[ Ans.\quad2x+\frac{1-x}{x^{2}+4}+\frac{3x}{x^{2}+16} \]

32.18. Find the partial fraction decomposition of $ $ .

\[ Ans.\quad\frac{x+1}{x^{2}+1}+\frac{3x-2}{(x^{2}+1)^{3}} \]

32.19. Find the partial fraction decomposition of $ $ .

\[ Ans.\quad x^{2}-x+\frac{x-2}{x^{2}+1}+\frac{3x-1}{(x^{2}+1)^{2}} \]

32.20. Find the partial fraction decomposition of $ $ .

\[ Ans.\quad\frac{5}{x^{2}+4}+\frac{x+3}{(x^{2}+4)^{2}}+\frac{-2x}{(x^{2}+1)^{2}} \]

32.21. Find the partial fraction decomposition of $ $ .

\[ Ans.\quad\frac{5}{x+1}+\frac{1-2x}{x^{2}-x+1} \]

32.22. Show that the partial fraction decomposition of $ $ can be written as $ + $ .