Partial Fraction Decomposition Integration Problems with Solutions: A Complete Tutorial
Partial fraction decomposition is one of the most used techniques for solving integrals involving rational functions. This method allows us to express a rational function as a sum of simpler fractions that can be integrated via other methods, such as substitution and completing the square. This tutorial will provide partial fraction decomposition integration problems with solutions.
The partial fraction decomposition problems given here will lead to systems that are best solved using techniques from linear algebra. Please see How to Solve a System of Equations Using Gaussian Elimination: A Step-by-Step Guide to learn one of these techniques.
This article is part of a three-part series on integrals of rational functions. The following infographic illustrates this topic.
Partial Fraction Decomposition
Partial fraction decomposition is a method for expressing a rational function as a sum of simpler fractions. This is useful in integration because then each fraction can be integrated separately. Specifically, this technique is useful when integrating proper rational functions where the degree of the numerator is less than the degree of the denominator, and the denominator consists of a reducible polynomial.
Steps for Integrating via Partial Fraction Decomposition
Step 1: Check that the Rational Function is Proper
To use partial fraction decomposition, the degree of the numerator must be less than that of the denominator. If it is not, you must first use polynomial long division. Please check out my article How to Integrate Using Polynomial Long Division with Examples to see how this is done.
Step 2: Factor the Denominator
Factor the denominator into linear or irreducible quadratic terms. If the denominator cannot be factored, skip to step 4.
Step 3: Find the Partial Fraction Decomposition
For help with this process, please refer to A Comprehensive Beginner’s Guide to Partial Fraction Decomposition, which explains how to find the partial fraction decomposition for a rational function.
Step 4: Integrate Each Term
Once the function is rewritten as a sum of simpler fractions, integrate each term separately using techniques discussed in the articles Basic Integration Problems for Beginners, The Ultimate Step-by-Step Guide to Solving Integrals Using Substitution, and The Ultimate Guide on How to Solve Integrals by Completing the Square.
Worked Out Example
Example 1: Evaluate:
(a) \( \int \frac{50}{x^2(x^2 + 6x + 10)} dx \)
(b) \( \int \frac{100}{x(x^2 + 6x + 10)^2} dx \)
Solution: (a) \( \int \frac{50}{x^2(x^2 + 6x + 10)} dx \)
Setting up the partial fraction decomposition, we get
$$\frac{50}{x^2(x^2 + 6x + 10)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 6x + 10}.$$
Next, multiplying both sides by \( x^2(x^2 + 6x + 10) \) gives
$$50 = Ax(x^2 + 6x + 10) + B(x^2 + 6x + 10) + (Cx + D)x^2.$$
By expanding the right-hand side of the equation, we obtain
$$50 = Ax^3 + 6Ax^2 + 10Ax + Bx^2 + 6Bx + 10B + Cx^3 + Dx^2.$$
Adding like terms, we find
$$50 = (A + C)x^3 + (6A+ B + D)x^2 + (10A + 6B)x + 10B.$$
Next, we equate coefficients to obtain the system of equations
$$\begin{aligned}
A + C &= 0 \\
6A + B + D &= 0 \\
10A + 6B &= 0 \\
10B &= 50
\end{aligned}.$$
The solution to this system is
$$A = -3, B = 5, C = 3, D = 13.$$
Thus,
$$\frac{50}{x^2(x^2 + 6x + 10)} = -\frac{3}{x} + \frac{5}{x^2} + \frac{3x + 13}{x^2 + 6x + 10}.$$
Our integral is then
$$\int -\frac{3}{x} + \frac{5}{x^2} + \frac{3x + 13}{x^2 + 6x + 10} dx.$$
This is equivalent to
$$\int -\frac{3}{x} + \frac{5}{x^2} + \frac{3}{2}\frac{2x + 6}{x^2 + 6x + 10} + \frac{4}{x^2 + 6x + 10} dx.$$
Next, use linearity to split the integral into three integrals as follows:
$$\int \frac{-3}{x} + \frac{5}{x^2} dx + \frac{3}{2} \int \frac{2x + 6}{x^2 + 6x + 10} dx + 4\int \frac{1}{x^2 + 6x + 10} dx.$$
Now, we complete the square in the third integral. To do so, we begin by rewriting 10 as 9 + 1, which gives
$$\int \frac{-3}{x} + \frac{5}{x^2} dx + \frac{3}{2} \int \frac{2x + 6}{x^2 + 6x + 10} dx + 4\int \frac{1}{x^2 + 6x + 9 + 1} dx.$$
Since, \( x^2 + 6x + 9 = (x + 3)^2 \), this is
$$\int \frac{-3}{x} + \frac{5}{x^2} dx + \frac{3}{2} \int \frac{2x + 6}{x^2 + 6x + 10} dx + 4\int \frac{1}{(x + 3)^2 + 1} dx.$$
In the second integral, let \( u = x^2 + 6x + 10 \), then \( du = (2x + 6)dx \). For the third integral, let \( v = x + 3 \), then \( dv = dx \). Substituting into the integrals, we have
$$\int \frac{-3}{x} + \frac{5}{x^2} dx + \frac{3}{2} \int \frac{1}{u} du + 4\int \frac{1}{v^2 + 1} dv.$$
This is equivalent to
$$\int \frac{-3}{x} + 5x^{-2} dx + \frac{3}{2} \int \frac{1}{u} du + 4\int \frac{1}{v^2 + 1}.$$
Integrating gives
$$-3\ln{|x|} – 5x^{-1} + \frac{3}{2}\ln{|u|} + 4\arctan{v} + C.$$
In terms of positive exponents, this is
$$-3\ln{|x|} – \frac{5}{x} + \frac{3}{2}\ln{|u|} + 4\arctan{v} + C.$$
Back-substituting gives us a final answer of
$$-3\ln{|x|} – \frac{5}{x} + \frac{3}{2}\ln{|x^2 + 6x + 10|} + 4\arctan{(x + 3)} + C.$$
(b) \( \int \frac{100}{x(x^2 + 6x + 10)^2} dx \)
Setting up the partial fraction decomposition, we get
$$\frac{100}{x(x^2 + 6x + 10)^2} = \frac{A}{x} + \frac{Bx + C}{x^2 + 6x + 10} + \frac{Dx + E}{(x^2 + 6x + 10)^2}.$$
Next, multiplying both sides by \( x(x^2 + 6x + 10)^2 \) gives
$$100 = A(x^2 + 6x + 10)^2 + (Bx+C)x(x^2 + 6x + 10) + (Dx + E)x.$$
By expanding the right-hand side of the equation, we obtain
$$100 = A(x^4 + 12x^3 + 56x^2 + 120x + 100) + (Bx + C)(x^3 + 6x^2 + 10x) + (Dx + E)x.$$
Distributing we get
$$100 = Ax^4 + 12Ax^3 + 56Ax^2 + 120Ax + 100A + Bx^4 + 6Bx^3 + 10Bx^2 + Cx^3 + 6Cx^2 + 10Cx + Dx^2 + Ex.$$
Adding like terms, we find
$$100 = (A + B)x^4 + (12A + 6B + C)x^3 + (56A + 10B + 6C + D)x^2 + (120A + 10C + E)x + 100A.$$
Next, we equate coefficients to obtain the system of equations
$$\begin{aligned}
A + B &= 0 \\
12A + 6B + C &= 0 \\
56A + 10B + 6C + D &= 0 \\
120A + 10C + E &= 0 \\
100A &= 100
\end{aligned}.$$
The solution to this system is
$$A = 1, B = -1, C = -6, D = -10, E = -60.$$
Thus,
$$\frac{100}{x(x^2 + 6x + 10)^2} = \frac{1}{x} – \frac{x + 6}{x^2 + 6x + 10} – \frac{10x + 60}{(x^2 + 6x + 10)^2}.$$
Our integral is then
$$\int \frac{1}{x} – \frac{x + 6}{x^2 + 6x + 10} – \frac{10x + 60}{(x^2 + 6x + 10)^2} dx,$$
which is equivalent to
$$\int \frac{1}{x} – \frac{1}{2}\frac{2x + 6}{x^2 + 6x + 10} – \frac{3}{x^2 + 6x + 10} – 5\frac{2x + 6}{(x^2 + 6x + 10)^2} – \frac{30}{(x^2 + 6x + 10)^2} dx.$$
Next, we use linearity to split the integral into four integrals as follows:
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{2x + 6}{x^2 + 6x + 10} + 5\frac{2x + 6}{(x^2 + 6x + 10)^2} dx – 3\int \frac{1}{x^2 + 6x +10} dx – 30\int \frac{1}{(x^2 + 6x + 10)^2} dx.$$
Now, we complete the square in the third and fourth integrals. To begin, we write 10 as 9 + 1, which gives
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{2x + 6}{x^2 + 6x + 10} + 5\frac{2x + 6}{(x^2 + 6x + 10)^2} dx – 3\int \frac{1}{x^2 + 6x + 9 + 1} dx – 30\int \frac{1}{(x^2 + 6x + 9 + 1)^2} dx.$$
This is equivalent to
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{2x + 6}{x^2 + 6x + 10} + 5\frac{2x + 6}{(x^2 + 6x + 10)^2} dx – 3\int \frac{1}{(x + 3)^2 + 1} dx – 30\int \frac{1}{((x + 3)^2 + 1)^2} dx.$$
In the second integral, let \( u = x^2 + 6x + 10 \), then \( du = (2x + 6)dx \). For the third integral, let \( v = x + 3 \), then \( dv = dx \). In the fourth integral let \( x + 3 = \tan{\theta} \), then \( dx = \sec^2{\theta} d\theta \). Substituting into the integrals, we have
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{1}{u} + 5\frac{1}{u^2} du – 3\int \frac{1}{v^2 + 1} dv – 30\int \frac{1}{(\tan^2{\theta} + 1)^2}\sec^2{\theta} d\theta.$$
Applying the Pythagorean identity in the fourth integral gives
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{1}{u} + 5\frac{1}{u^2} du – 3\int \frac{1}{v^2 + 1} dv – 30\int \frac{1}{(\sec^2{\theta})^2}\sec^2{\theta} d\theta.$$
Simplifying the fourth integral, we get
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{1}{u} + 5\frac{1}{u^2} du – 3\int \frac{1}{v^2 + 1} dv – 30\int \frac{1}{\sec^2{\theta}} d\theta.$$
Applying exponential properties and the reciprocal identity, this is
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{1}{u} + 5u^{-2} du – 3\int \frac{1}{v^2 + 1} dv – 30\int \cos^2{\theta} d\theta.$$
By the power-reducing identity, this is equivalent to
$$\int \frac{1}{x} dx – \int \frac{1}{2}\frac{1}{u} + 5u^{-2} du – 3\int \frac{1}{v^2 + 1} dv – 15\int 1 + \cos{(2\theta)} d\theta.$$
Integrating we get
$$\ln{|x|} – \frac{1}{2}\ln{|u|} + 5u^{-1} – 3\arctan{v} – 15\theta – \frac{15}{2}\sin{(2\theta)} + C.$$
In terms of positive exponents, this is
$$\ln{|x|} – \frac{1}{2}\ln{|u|} + \frac{5}{u} – 3\arctan{v} – 15\theta – 15\sin{\theta}\cos{\theta} + C.$$
Shown below is the right triangle for our substitution.
Using the triangle and back-substituting gives us
$$\ln{|x|} – \frac{1}{2}\ln{|x^2 + 6x + 10|} + \frac{5}{x^2 + 6x + 10} – 3\arctan{(x + 3)} – 15\arctan{(x + 3)} – \frac{15(x + 3)}{(x + 3)^2 + 1} + C.$$
This is equivalent to
$$\ln{|x|} – \frac{1}{2}\ln{|x^2 + 6x + 10|} + \frac{5}{x^2 + 6x + 10} – 18\arctan{(x + 3)} – \frac{15(x + 3)}{x^2 + 6x + 10} + C.$$
Simplifying, we arrive at a final answer of
$$\ln{|x|} – \frac{1}{2}\ln{|x^2 + 6x + 10|} – 18\arctan{(x + 3)} – \frac{15x + 40}{x^2 + 6x + 10} + C.$$
Conclusion
Partial fraction decomposition is a powerful method used to integrate rational functions. With it, along with polynomial long division and completing the square, you can, in theory, integrate any rational function. This guide provided partial fraction decomposition integration problems with solutions to help you integrate rational functions.
Further Reading
Examples of Rationalizing Substitution in Calculus: A Beginner-Friendly Guide – Several substitutions exist for transforming difficult integrals into rational functions. Rationalizing substitution is one of them.
