# Partial Fractions Practice Problems with Answers

> 12 partial fraction decomposition practice problems with answers: distinct and repeated factors, irreducible quadratics and long division first.

Source: https://practicecalculus.com/practice-problems/partial-fractions
Updated: 2026-11-06

$$ \frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2 and AP Calculus BC
- **You need:** Factoring, polynomial long division and $\int\frac{1}{x}\,dx = \ln\lvert x\rvert + C$
- **Main idea:** Split one hard fraction into several easy ones, then integrate each
- **Practice with new numbers:** [Partial fractions lesson](/learn/calculus-2/integration-techniques/partial-fractions)

## Key takeaways

- If the top degree is at least the bottom degree, do long division first.
- Give each linear factor its own constant, each repeated factor one term per power, and each irreducible quadratic a $Bx + C$ numerator.
- Find constants fast by plugging in the roots of the denominator, then match coefficients for anything left.

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:

| Factor | Terms to include |
| --- | --- |
| $(x - a)$ | $\dfrac{A}{x - a}$ |
| $(x - a)^2$ | $\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$ |
| $x^2 + bx + c$ (no real roots) | $\dfrac{Bx + C}{x^2 + bx + c}$ |

For worked examples and the reasoning, read the [partial fractions guide](/blog/partial-fractions).

## 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

| Problem | Answer |
| --- | --- |
| 1 | $\ln\lvert x\rvert - \ln\lvert x + 1\rvert + C$ |
| 2 | $\dfrac{1}{4}\ln\left\lvert\dfrac{x - 2}{x + 2}\right\rvert + C$ |
| 3 | $2\ln\lvert x + 1\rvert + 3\ln\lvert x - 2\rvert + C$ |
| 4 | $2\ln\lvert x - 1\rvert - \ln\lvert x + 2\rvert + C$ |
| 5 | $-\ln\lvert x\rvert - \dfrac{1}{x} + \ln\lvert x + 1\rvert + C$ |
| 6 | $3\ln\lvert x + 1\rvert - \dfrac{2}{x + 1} + C$ |
| 7 | $\ln\lvert x\rvert - \dfrac{1}{2}\ln(x^2 + 1) + C$ |
| 8 | $\dfrac{1}{2}\ln\lvert x + 1\rvert + \dfrac{1}{4}\ln(x^2 + 1) + \dfrac{1}{2}\arctan x + C$ |
| 9 | $\dfrac{x^2}{2} + x + 2\ln\lvert x - 1\rvert + C$ |
| 10 | $\dfrac{x^2}{2} + \dfrac{1}{2}\ln\lvert x^2 - 1\rvert + C$ |
| 11 | $\dfrac{1}{2}\ln\dfrac{3}{2} \approx 0.2027$ |
| 12 | $2\ln\dfrac{4}{3} \approx 0.5754$ |

## 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 $x = -1$: $B = -1$:

$$\int\left(\frac{1}{x} - \frac{1}{x + 1}\right)dx = \ln|x| - \ln|x + 1| + C$$

**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 $A = -1$:

$$\int\left(-\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x + 1}\right)dx = -\ln|x| - \frac{1}{x} + \ln|x + 1| + C$$

**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:

$$\int\left(\frac{1/2}{x + 1} + \frac{1}{2}\cdot\frac{x}{x^2 + 1} + \frac{1}{2}\cdot\frac{1}{x^2 + 1}\right)dx$$

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, $\frac{1}{x^2 - 1} = \frac{1}{2}\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)$:

$$\frac{1}{2}\left[\ln\frac{x - 1}{x + 1}\right]_2^3 = \frac{1}{2}\left(\ln\frac{1}{2} - \ln\frac{1}{3}\right) = \frac{1}{2}\ln\frac{3}{2}$$

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

$$2\Big[\ln(x + 1) - \ln(x + 2)\Big]_0^1 = 2\big[(\ln 2 - \ln 3) - (0 - \ln 2)\big] = 2\ln\frac{4}{3}$$

## 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](/learn/calculus-2/integration-techniques/partial-fractions) and the [repeated factors lesson](/learn/calculus-2/integration-techniques/partial-fractions-repeated). See also the [integral of 1/(x² − 1)](/integrals/1-over-x-squared-minus-1). Then put every method together in the [techniques practice set](/learn/calculus-2/integration-techniques/techniques-practice).

## Frequently asked questions

### When do you use partial fractions?

Use it for a rational function, a polynomial over a polynomial, when the bottom factors and no simple u-substitution works. If the top is the derivative of the bottom, as in $\frac{2x}{x^2 + 1}$, use u-substitution instead.

### What is the cover-up method?

For a distinct linear factor $(x - a)$, cover it in the original fraction and plug $x = a$ into what remains. The result is the constant over $(x - a)$. It is the same as plugging in the root after clearing denominators.

### How do you handle a repeated factor like (x+1)²?

Include one term for each power: $\frac{A}{x + 1} + \frac{B}{(x + 1)^2}$. The squared term integrates with the power rule, giving $-\frac{B}{x + 1}$, not a log. Problems 5 and 6 show this.