Partial Fractions Practice Problems with Answers

These partial fractions practice problems start with two distinct linear factors and move on to repeated factors, irreducible quadratics, long division and definite integrals. Try each one first. The answers come first so you can check quickly, then the full solutions.

The setup for each type of factor in the denominator:

For worked examples and the reasoning, read the partial fractions guide.

Problems

Distinct linear factors

1. \displaystyle\int\frac{1}{x(x + 1)}\,dx
2. \displaystyle\int\frac{1}{x^2 - 4}\,dx
3. \displaystyle\int\frac{5x - 1}{(x + 1)(x - 2)}\,dx
4. \displaystyle\int\frac{x + 5}{x^2 + x - 2}\,dx

Repeated factors

5. \displaystyle\int\frac{1}{x^2(x + 1)}\,dx
6. \displaystyle\int\frac{3x + 5}{(x + 1)^2}\,dx

Irreducible quadratics

7. \displaystyle\int\frac{1}{x(x^2 + 1)}\,dx
8. \displaystyle\int\frac{x^2 + x + 1}{(x + 1)(x^2 + 1)}\,dx

Long division first

9. \displaystyle\int\frac{x^2 + 1}{x - 1}\,dx
10. \displaystyle\int\frac{x^3}{x^2 - 1}\,dx

Definite integrals

11. \displaystyle\int_2^3\frac{1}{x^2 - 1}\,dx
12. \displaystyle\int_0^1\frac{2}{(x + 1)(x + 2)}\,dx

Answers

Step-by-step solutions

1. Write \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x + 1}, so 1 = A(x + 1) + Bx. Plug in x = 0: A = 1. Plug in x = -1: B = -1:

2. Factor: x^2 - 4 = (x - 2)(x + 2). Then 1 = A(x + 2) + B(x - 2). At x = 2, A = \frac{1}{4}. At x = -2, B = -\frac{1}{4}. Combine the two logs into one.

3. 5x - 1 = A(x - 2) + B(x + 1). At x = -1: -6 = -3A, so A = 2. At x = 2: 9 = 3B, so B = 3.

4. Factor: x^2 + x - 2 = (x + 2)(x - 1). Then x + 5 = A(x - 1) + B(x + 2). At x = 1: 6 = 3B, so B = 2. At x = -2: 3 = -3A, so A = -1.

5. The factor x^2 is repeated, so write \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}. Then 1 = Ax(x + 1) + B(x + 1) + Cx^2. At x = 0: B = 1. At x = -1: C = 1. Match x^2 terms: A + C = 0, so A = -1:

6. Write the top in terms of x + 1: 3x + 5 = 3(x + 1) + 2. So the fraction is \frac{3}{x + 1} + \frac{2}{(x + 1)^2}. The second term integrates with the power rule to -\frac{2}{x + 1}.

7. Write \frac{A}{x} + \frac{Bx + C}{x^2 + 1}, so 1 = A(x^2 + 1) + (Bx + C)x. At x = 0: A = 1. Match x^2: A + B = 0, so B = -1. Match x: C = 0. The term \frac{-x}{x^2 + 1} integrates to -\frac{1}{2}\ln(x^2 + 1).

8. Write \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}, so x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 1). At x = -1: 1 = 2A, so A = \frac{1}{2}. Match x^2: 1 = A + B, so B = \frac{1}{2}. Match constants: 1 = A + C, so C = \frac{1}{2}. Then:

The three pieces give a log, a log by u-substitution and an arctangent.

9. The top has degree 2 and the bottom degree 1, so divide: x^2 + 1 = (x - 1)(x + 1) + 2. So the integrand is x + 1 + \frac{2}{x - 1}.

10. Divide: x^3 = x(x^2 - 1) + x, so the integrand is x + \frac{x}{x^2 - 1}. The second term is a u-substitution with u = x^2 - 1, giving \frac{1}{2}\ln|x^2 - 1|.

11. As in problem 2, \frac{1}{x^2 - 1} = \frac{1}{2}\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right):

12. \frac{2}{(x + 1)(x + 2)} = \frac{2}{x + 1} - \frac{2}{x + 2}:

Common mistakes on these problems

• Skipping long division. In problem 9, splitting \frac{x^2 + 1}{x - 1} without dividing first gives a setup that has no solution.
• One term for a repeated factor. \frac{1}{x^2(x + 1)} needs both \frac{A}{x} and \frac{B}{x^2}.
• A constant over a quadratic. An irreducible quadratic needs Bx + C on top, not just B.
• Integrating \frac{1}{(x + 1)^2} as a log. It is a power: \int(x + 1)^{-2}\,dx = -\frac{1}{x + 1}.

Keep practicing

For unlimited problems with instant feedback, try the partial fractions lesson and the repeated factors lesson. See also the integral of 1/(x² − 1). Then put every method together in the techniques practice set.