Decompose Rational Functions into Partial Fractions
Partial fraction decomposition is a method to break down complex rational expressions into simpler fractions. This calculator helps you decompose rational functions of the form P(x)/Q(x) where the degree of P is less than the degree of Q.
Enter Your Rational Function
Example Input
Numerator: 3x^2 + 2x - 1
Denominator: (x+1)(x-2)^2
Or: x^3 - 3x^2 + 4
Partial Fraction Result
Enter a rational function and click “Calculate” to see the partial fraction decomposition.
How Partial Fractions Work
- Factor the denominator Q(x) completely
- For each distinct linear factor (ax+b), include A/(ax+b)
- For repeated linear factors (ax+b)^n, include A?/(ax+b) + A?/(ax+b)� + … + A?/(ax+b)?
- For irreducible quadratics, include (Ax+B)/(quadratic)
- Solve for the unknown constants
A partial fraction calculator helps you split a rational function into a sum of simpler fractions that are easier to integrate, invert with Laplace transforms, or evaluate symbolically. This guide explains the logic behind partial fraction decomposition so you can understand what a calculator is doing, verify results by hand, and avoid common mistakes.
When Partial Fractions Apply
You can decompose a rational function P(x)Q(x)\frac{P(x)}{Q(x)}Q(x)P(x)? when:
- P(x)P(x)P(x) and Q(x)Q(x)Q(x) are polynomials, and
- deg?P(x)<deg?Q(x)\deg P(x) < \deg Q(x)degP(x)<degQ(x) (if not, perform polynomial long division first), and
- Q(x)Q(x)Q(x) factors over the reals into linear factors (x?a)(x-a)(x?a) and/or irreducible quadratics (x2+bx+c,?b2?4c<0)(x^2+bx+c, \, b^2-4c<0)(x2+bx+c,b2?4c<0).
Standard Decomposition Forms
- Distinct linear factors
If Q(x)=(x?a1)(x?a2)?(x?an)Q(x) = (x-a_1)(x-a_2)\cdots(x-a_n)Q(x)=(x?a1?)(x?a2?)?(x?an?), then P(x)Q(x)=?k=1nAkx?ak.\frac{P(x)}{Q(x)} = \sum_{k=1}^n \frac{A_k}{x-a_k}.Q(x)P(x)?=k=1?n?x?ak?Ak??. - Repeated linear factor (x?a)m(x-a)^m(x?a)m P(x)(x?a)m=A1x?a+A2(x?a)2+?+Am(x?a)m.\frac{P(x)}{(x-a)^m} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_m}{(x-a)^m}.(x?a)mP(x)?=x?aA1??+(x?a)2A2??+?+(x?a)mAm??.
- Irreducible quadratic (x2+bx+c)(x^2+bx+c)(x2+bx+c) P(x)x2+bx+c=Ax+Bx2+bx+c.\frac{P(x)}{x^2+bx+c} = \frac{Ax+B}{x^2+bx+c}.x2+bx+cP(x)?=x2+bx+cAx+B?.
- Repeated irreducible quadratic (x2+bx+c)m(x^2+bx+c)^m(x2+bx+c)m P(x)(x2+bx+c)m=?k=1mAkx+Bk(x2+bx+c)k.\frac{P(x)}{(x^2+bx+c)^m} = \sum_{k=1}^{m} \frac{A_k x + B_k}{(x^2+bx+c)^k}.(x2+bx+c)mP(x)?=k=1?m?(x2+bx+c)kAk?x+Bk??.
The Workflow (What a Calculator Does)
- Check improper vs proper.
If deg?P?deg?Q\deg P \ge \deg QdegP?degQ, perform long division: PQ=S(x)+R(x)Q(x) \frac{P}{Q} = S(x) + \frac{R(x)}{Q(x)}QP?=S(x)+Q(x)R(x)? where deg?R<deg?Q\deg R < \deg QdegR<degQ. - Factor the denominator.
Find linear and irreducible quadratic factors (with multiplicities). - Write the decomposition template.
Use the forms above to set unknown coefficients A,B,�A, B,\ldotsA,B,�. - Solve for coefficients.
- Multiply both sides by Q(x)Q(x)Q(x), expand, and equate coefficients of like powers of xxx; or
- Use the Heaviside (cover-up) method for distinct linear factors to find simple residues quickly; then finish with equating coefficients if needed.
- Verify.
Combine the found terms over a common denominator to confirm you recover P(x)/Q(x)P(x)/Q(x)P(x)/Q(x).
Worked Examples
1) Distinct linear factors
Decompose 3x+5×2?1=3x+5(x?1)(x+1)=Ax?1+Bx+1.\frac{3x+5}{x^2-1}=\frac{3x+5}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.x2?13x+5?=(x?1)(x+1)3x+5?=x?1A?+x+1B?.
Multiply by (x?1)(x+1)(x-1)(x+1)(x?1)(x+1): 3x+5=A(x+1)+B(x?1)=(A+B)x+(A?B).3x+5 = A(x+1) + B(x-1) = (A+B)x + (A-B).3x+5=A(x+1)+B(x?1)=(A+B)x+(A?B).
Equate coefficients: {A+B=3A?B=5?A=4,??B=?1.\begin{cases} A+B = 3\\ A-B = 5 \end{cases} \Rightarrow A=4,\; B=-1.{A+B=3A?B=5??A=4,B=?1.
So 3x+5×2?1=4x?1?1x+1.\frac{3x+5}{x^2-1} = \frac{4}{x-1} – \frac{1}{x+1}.x2?13x+5?=x?14??x+11?.
Cover-up check:
At x=1x=1x=1: A=3x+5x+1?x=1=82=4.\displaystyle A=\left.\frac{3x+5}{x+1}\right|_{x=1}=\frac{8}{2}=4.A=x+13x+5??x=1?=28?=4.
At x=?1x=-1x=?1: B=3x+5x?1?x=?1=2?2=?1.\displaystyle B=\left.\frac{3x+5}{x-1}\right|_{x=-1}=\frac{2}{-2}=-1.B=x?13x+5??x=?1?=?22?=?1.
2) Repeated linear factor
Decompose 2x+1(x?1)2=Ax?1+B(x?1)2.\frac{2x+1}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}.(x?1)22x+1?=x?1A?+(x?1)2B?.
Multiply by (x?1)2(x-1)^2(x?1)2: 2x+1=A(x?1)+B.2x+1 = A(x-1) + B.2x+1=A(x?1)+B.
Set x=1x=1x=1: 2(1)+1=3=B?B=32(1)+1=3=B\Rightarrow B=32(1)+1=3=B?B=3.
Then compare coefficients: 2x+1=Ax?A+32x+1 = A x – A + 32x+1=Ax?A+3.
Thus A=2A=2A=2 and ?A+3=1-A+3=1?A+3=1 checks.
So 2x+1(x?1)2=2x?1+3(x?1)2.\frac{2x+1}{(x-1)^2}=\frac{2}{x-1}+\frac{3}{(x-1)^2}.(x?1)22x+1?=x?12?+(x?1)23?.
3) Irreducible quadratic factor
Decompose 2×2+3x+1(x+1)(x2+1)=Ax+1+Bx+Cx2+1.\frac{2x^2+3x+1}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}.(x+1)(x2+1)2×2+3x+1?=x+1A?+x2+1Bx+C?.
Multiply by (x+1)(x2+1)(x+1)(x^2+1)(x+1)(x2+1): 2×2+3x+1=A(x2+1)+(Bx+C)(x+1).2x^2+3x+1 = A(x^2+1) + (Bx+C)(x+1).2×2+3x+1=A(x2+1)+(Bx+C)(x+1).
Expand RHS: Ax2+A+Bx2+Bx+Cx+C=(A+B)x2+(B+C)x+(A+C).A x^2 + A + Bx^2 + Bx + Cx + C = (A+B)x^2 + (B+C)x + (A+C).Ax2+A+Bx2+Bx+Cx+C=(A+B)x2+(B+C)x+(A+C).
Match coefficients: {A+B=2B+C=3A+C=1?A=0,??B=2,??C=1.\begin{cases} A+B = 2\\ B+C = 3\\ A+C = 1 \end{cases} \Rightarrow A=0,\; B=2,\; C=1.????A+B=2B+C=3A+C=1??A=0,B=2,C=1.
So 2×2+3x+1(x+1)(x2+1)=0x+1+2x+1×2+1=2x+1×2+1.\frac{2x^2+3x+1}{(x+1)(x^2+1)}=\frac{0}{x+1}+\frac{2x+1}{x^2+1}=\frac{2x+1}{x^2+1}.(x+1)(x2+1)2×2+3x+1?=x+10?+x2+12x+1?=x2+12x+1?.
(Here the linear term ends up capturing everything; this is a good sanity check example.)
Long Division First: An Improper Example
x3+2×2+1×2?1\frac{x^3+2x^2+1}{x^2-1}x2?1×3+2×2+1?
Divide: x3+2×2+1=(x2?1)(x+2)+(2x+3)x^3+2x^2+1 = (x^2-1)(x+2) + (2x+3)x3+2×2+1=(x2?1)(x+2)+(2x+3).
Thus x3+2×2+1×2?1=x+2+2x+3×2?1=x+2+Ax?1+Bx+1.\frac{x^3+2x^2+1}{x^2-1} = x+2 + \frac{2x+3}{x^2-1} = x+2 + \frac{A}{x-1}+\frac{B}{x+1}.x2?1×3+2×2+1?=x+2+x2?12x+3?=x+2+x?1A?+x+1B?.
Proceed as in Example 1 to find AAA and BBB.
Cover-Up Method (Heaviside) in a Nutshell
For distinct linear factors: P(x)?k(x?ak)=?kAkx?ak,Ak=P(x)?j?k(x?aj)?x=ak.\frac{P(x)}{\prod_{k}(x-a_k)}=\sum_k \frac{A_k}{x-a_k},\quad A_k=\left.\frac{P(x)}{\prod_{j\ne k}(x-a_j)}\right|_{x=a_k}.?k?(x?ak?)P(x)?=k??x?ak?Ak??,Ak?=?j?=k?(x?aj?)P(x)??x=ak??.
This gives the simple residues quickly. For repeated factors and quadratics, use equating coefficients.
Common Mistakes to Avoid
- Skipping long division when the fraction is improper.
- Using A/(x2+bx+c)A/(x^2+bx+c)A/(x2+bx+c) for an irreducible quadratic; it must be (Ax+B)/(x2+bx+c)(Ax+B)/(x^2+bx+c)(Ax+B)/(x2+bx+c).
- Forgetting terms for repeated factors (you need a term for each power).
- Not verifying by recombining over the common denominator.
Applications
- Integration: Decompose first, then integrate term by term.
- Differential Equations & Laplace Transforms: Convert rational expressions to simpler forms for inverse transforms.
- Algebraic Simplification: Make expressions easier to manipulate or evaluate.
FAQ
Do I always need to factor completely?
Yes, decompose with respect to linear and irreducible quadratic factors. Over the reals, not all quadratics factor into linears.
What if factoring is hard?
Symbolic calculators use algorithms (e.g., square-free factorization). By hand, try rational root tests and complete the square for quadratics.
How do I check my decomposition?
Combine your result over a common denominator and confirm you recover the original numerator.
Can I use partial fractions for complex numbers?
Yes. Over C\mathbb{C}C, every polynomial factors into linear terms; coefficients may be complex.
Why do some coefficients become zero?
It simply means certain partial terms are not needed after solving the system�this is normal.
Conclusion
A partial fraction calculator automates factorization, template setup, coefficient solving, and verification. Understanding the process ensures you can trust the output, spot mistakes, and handle edge cases�especially when integrating or applying Laplace transforms.
