1.25
-1.25
2
-3
0
5
1.25
-1.25
2
-3
0
5
The Partial Fraction Decomposition Calculator breaks down a rational expression of the form $$\frac{N}{(x - r_1)(x - r_2)}$$ into its partial fraction components $$\frac{A}{x - r_1} + \frac{B}{x - r_2}$$. Partial fraction decomposition is a fundamental technique in algebra and calculus that rewrites a complicated rational expression as a sum of simpler fractions, making integration, inverse Laplace transforms, and algebraic simplification vastly more manageable.
The method is most commonly encountered in integral calculus, where integrating $$\frac{N}{(x - r_1)(x - r_2)}$$ directly is difficult, but integrating $$\frac{A}{x - r_1} + \frac{B}{x - r_2}$$ is straightforward — each term integrates to a natural logarithm. This technique is also essential in differential equations (particularly when using Laplace transforms), control theory (transfer function analysis), signal processing, and discrete mathematics (solving linear recurrence relations via generating functions).
The core idea is that any proper rational function whose denominator factors into distinct linear factors can be decomposed into a sum of fractions, each with one linear factor in its denominator and a constant in its numerator. For the case of two distinct linear factors, we write: $$\frac{N}{(x - r_1)(x - r_2)} = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$
To find $$A$$ and $$B$$, multiply both sides by $$(x - r_1)(x - r_2)$$ to get $$N = A(x - r_2) + B(x - r_1)$$. Setting $$x = r_1$$ eliminates the $$B$$ term: $$N = A(r_1 - r_2)$$, so $$A = \frac{N}{r_1 - r_2}$$. Setting $$x = r_2$$ eliminates the $$A$$ term: $$N = B(r_2 - r_1)$$, so $$B = \frac{N}{r_2 - r_1} = \frac{-N}{r_1 - r_2}$$. Note that $$A + B = 0$$ when the numerator is a constant, which serves as a built-in verification check.
This calculator handles the case of a constant numerator over a product of two distinct linear factors — the most common scenario in introductory calculus courses. You enter the numerator constant $$N$$ and the two roots $$r_1$$ and $$r_2$$ of the denominator, and the calculator computes the coefficients $$A$$ and $$B$$, the expanded denominator form, and verification sums to confirm correctness. The roots must be distinct ($$r_1 \neq r_2$$); if they are equal, the decomposition requires a different form involving $$(x - r)^2$$.
Understanding partial fractions opens the door to powerful integration techniques and is a prerequisite for advanced topics in engineering mathematics. Whether you are a calculus student working through integration exercises, an engineering student computing inverse transforms, or a professional needing quick decomposition results, this calculator delivers accurate answers with complete transparency into the solution process.
Enter the numerator constant $$N$$ and the two roots $$r_1$$ and $$r_2$$. The calculator decomposes:
$$\frac{N}{(x - r_1)(x - r_2)} = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$
Using the cover-up method:
The denominator is also expanded: $$(x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$$, and verification checks confirm $$A + B = 0$$ and $$-Ar_2 - Br_1 = N$$.
The coefficients A and B are the numerators of the partial fractions. The decomposition means: to work with $$\frac{N}{(x-r_1)(x-r_2)}$$, you can instead work with $$\frac{A}{x-r_1} + \frac{B}{x-r_2}$$. For integration, this means $$\int \frac{N}{(x-r_1)(x-r_2)}dx = A\ln|x-r_1| + B\ln|x-r_2| + C$$. The verification fields confirm correctness: A+B should equal 0 for a constant numerator, and the constant verification should equal N.
Inputs
Results
A = 5/(1−(−3)) = 5/4 = 1.25 and B = −5/4 = −1.25. So 5/((x−1)(x+3)) = 1.25/(x−1) − 1.25/(x+3). Verify: A+B = 0 and −A(−3) − B(1) = 3.75 + 1.25 = 5.
Inputs
Results
A = 1/(2−5) = −1/3 and B = 1/(5−2) = 1/3. So 1/((x−2)(x−5)) = (−1/3)/(x−2) + (1/3)/(x−5). Verify: A+B = 0 and −(−1/3)(5) − (1/3)(2) = 5/3 − 2/3 = 1.
Partial fraction decomposition is a technique that rewrites a rational expression (a fraction of polynomials) as a sum of simpler fractions. Each simpler fraction has a denominator that is one of the factors of the original denominator, making subsequent operations like integration much easier.
This calculator handles the case of two distinct linear factors. When the roots are equal (repeated root), the decomposition takes a different form: $$\frac{N}{(x-r)^2} = \frac{A}{x-r} + \frac{B}{(x-r)^2}$$, which requires a different computation method.
After decomposing $$\frac{N}{(x-r_1)(x-r_2)}$$ into $$\frac{A}{x-r_1} + \frac{B}{x-r_2}$$, each term integrates directly: $$\int \frac{A}{x-r_1}dx = A\ln|x-r_1| + C$$. Without decomposition, the integral would require a much more complex approach.
The cover-up method (Heaviside's method) finds partial fraction coefficients by strategically substituting the root values. To find $$A$$, "cover up" the factor $$(x-r_1)$$ in the original expression and substitute $$x = r_1$$. This instantly gives $$A = \frac{N}{r_1 - r_2}$$. It is the fastest method for distinct linear factors.
Yes. For a denominator with three or more distinct linear factors, you write one fraction per factor. For irreducible quadratic factors, the numerator is a linear expression $$Ax + B$$ rather than a constant. The general technique extends to any factorable denominator.
When the numerator is a constant (degree 0) and the denominator is degree 2, the decomposition must reconstruct the original: $$A(x-r_2) + B(x-r_1) = (A+B)x + (-Ar_2 - Br_1)$$. Matching the $$x$$ coefficient to zero (since the numerator has no $$x$$ term) forces $$A + B = 0$$. This is a useful correctness check.
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