A Comprehensive Beginner’s Guide to Partial Fraction Decomposition

Step 3: Set Up the Partial Fraction Decomposition

For each factor in the denominator, set up a corresponding fraction in the partial fraction decomposition:

For distinct linear factors, the partial fraction decomposition takes the form

$$\frac{A}{ax + b}.$$
For repeated linear factors, the decomposition becomes

$$\sum_{i = 1}^n \frac{A_i}{(a_ix + b_i)^i}.$$

For irreducible quadratic factors, the decomposition will have the form

$$\frac{Ax + B}{ax^2 + bx + c}.$$

For repeated irreducible quadratic factors, the decomposition becomes

$$\sum_{i = 1}^n \frac{A_ix + B_i}{(a_ix^2 + b_ix + c_i)^i}.$$

Step 4: Solve for the Unknown Constants

Now, solve for the unknown constants in the partial fraction decomposition. To do this, multiply both sides of the equation by the common denominator to eliminate the fractions. Then, equate coefficients or substitute suitable values of x to find the unknowns.

Step 5: Finalize Result

Now that you’ve found all the coefficients, you can plug them into the decomposition from earlier. This is your final partial fraction decomposition.

Worked Out Examples

Example 1: Find the partial fraction decomposition of \( \frac{6}{(x + 1)(x + 2)} \).

Solution: Setting up the partial fraction decomposition, we get

$$\frac{6}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}.$$

Next, multiplying both sides by \( (x + 1)(x + 2) \) gives

$$6 = A(x + 2) + B(x + 1).$$

Now, we determine the coefficients by choosing convenient values of x:

$$x = -1 \Rightarrow 6 = A.$$

$$x = -2 \Rightarrow 6 = -B \Rightarrow -6 = B.$$

Thus,

$$\frac{6}{(x + 1)(x + 2)} = \frac{6}{x + 1} – \frac{6}{x + 2}.$$

Example 2: Find the partial fraction decomposition of \( \frac{x}{(x + 1)^2} \).

Solution: Setting up the partial fraction decomposition, we get

$$\frac{x}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2}.$$

Next, multiplying both sides by \( (x + 1)^2 \) gives

$$x = A(x + 1) + B.$$

By expanding the right-hand side of the equation, we obtain

$$x = Ax + A + B.$$

Next, we equate coefficients to obtain the system of equations

$$\begin{aligned}
A &= 1 \\
A + B &= 0
\end{aligned}.$$

Substituting A = 1 into the second equation gives

$$1 + B = 0 \Rightarrow B = -1.$$

Thus,

$$\frac{x}{(x + 1)^2} = \frac{1}{x + 1} – \frac{1}{(x + 1)^2}.$$

Example 3: Find the partial fraction decomposition of \( \frac{1}{(x^2 + 1)(x^2 + 2)} \).

Solution: Setting up the partial fraction decomposition, we get

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

Next, multiplying both sides by \( (x^2 + 1)(x^2 + 2) \) gives

$$1 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1).$$

By expanding the right-hand side of the equation, we obtain

$$1 = Ax^3 + Bx^2 + 2Ax + 2B + Cx^3 + Dx^2 + Cx + D.$$

Adding like terms, we find

$$1 = (A + C)x^3 + (B + D)x^2 + (2A + C)x + (2B + D).$$

Next, we equate coefficients to obtain the system of equations

$$\begin{aligned}
A + C &= 0 \\
B + D &= 0 \\
2A + C &= 0 \\
2B + D &= 1
\end{aligned}.$$

This is essentially 2 different systems of equations with 2 equations and 2 unknowns. We first consider

$$\begin{aligned}
A + C &= 0 \\
2A + C &= 0
\end{aligned}.$$

Multiplying the first equation by 2, this is equivalent to

$$\begin{aligned}
2A + 2C &= 0 \\
2A + C &= 0
\end{aligned}.$$

Subtracting the second equation from the first equation, we get

$$C = 0.$$

Substituting C = 0 into the first equation \( A + C = 0 \), we obtain

$$A = 0.$$

We now consider

$$\begin{aligned}
B + D &= 0 \\
2B + D &= 1
\end{aligned}.$$

Multiplying the first equation by 2, this is equivalent to

$$\begin{aligned}
2B + 2D &= 0 \\
2B + D&= 1
\end{aligned}.$$

Subtracting the second equation from the first equation, we get

$$D = -1.$$

Substituting D = -1 into the first equation \( B + D = 0 \), we obtain

$$B – 1 = 0 \Rightarrow B = 1.$$

Thus,

$$\frac{1}{(x^2 + 1)(x^2 + 2)} = \frac{1}{x^2 + 1} – \frac{1}{x^2 + 2}.$$

Example 4: Find the partial fraction decomposition of \( \frac{x^2}{(x^2 + 1)^2} \).

Solution: Setting up the partial fraction decomposition, we get

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

Next, multiplying both sides by \( (x^2 + 1)^2 \) gives

$$x^2 = (Ax + B)(x^2 + 1) + Cx + D.$$

By expanding the right-hand side of the equation, we obtain

$$x^2 = Ax^3 + Bx^2 + Ax + B + Cx + D.$$

Adding like terms, we find

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

Next, we equate coefficients to obtain the system of equations

$$\begin{aligned}
A &= 0 \\
B &= 1 \\
A + C &= 0 \\
B + D &= 0
\end{aligned}.$$

This is essentially two different systems of equations with two equations and two unknowns. We first consider

$$\begin{aligned}
A &= 0 \\
A + C &= 0
\end{aligned}.$$

Substituting A = 0 into the second equation, we obtain

$$C = 0.$$

We now consider

$$\begin{aligned}
B &= 1 \\
B + D &= 0
\end{aligned}.$$

Substituting B = 1 into the second equation, we obtain

$$1 + D = 0 \Rightarrow D = -1.$$

Thus,

$$\frac{x^2}{(x^2 + 1)^2} = \frac{1}{x^2 + 1} – \frac{1}{(x^2 + 1)^2}.$$

Conclusion

Partial fraction decomposition is a powerful technique for splitting rational expressions into simpler fractions. With the steps outlined in this guide, you now have the skills to find the partial fraction decomposition of rational functions with distinct linear, repeated linear, irreducible quadratic, and repeated irreducible quadratic factors in the denominator. Practicing these methods will enhance your algebra skills and prepare you for more advanced topics, such as integrating rational functions and finding inverse Laplace transforms.

Frequently Asked Questions