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The Partial Fraction Decomposition Calculator is an online tool that displays the expansion of polynomial rational functions. STUDYQUERIES’s online partial fraction decomposition calculator tool speeds up calculations and displays partial fraction expansions in a fraction of a second.

**How to Use the Partial Fraction Decomposition Calculator?**

To use the partial fraction decomposition calculator, follow these steps:

**Step 1:**Enter the numerator and denominator polynomials in the respective input fields**Step 2:**To obtain the expansion, click the button “Submit”**Step 3:**In the new window, the partial fraction decomposition of the given polynomial rational function will be displayed

Partial Fraction Decomposition Calculator

**What Is Partial Fraction Decomposition And Calculator?**

We examine the antiderivatives of rational functions in this article. Recall that rational functions are functions of the form $$f(x)=\frac{p(x)}{q(x)}$$

where \(p(x)\) and \(q(x)\) are polynomials and \(q(x)\neq 0\). There are many contexts in which such functions are used, including the solution of certain fundamental differential equations.

This section begins with an example that demonstrates the motivation behind it. Considering the integral

$$\int_{}^{}\frac{1}{x^2-1}dx$$

We do not have a simple formula for this (if the denominator were \((x^2+1)\), we would recognize the antiderivative as being the arctangent function). You can solve it using Trigonometric Substitution, but note how easy it is to evaluate the integral once you realize:

$$\frac{1}{x^{2}-1}=\frac{1/2}{x-1}-\frac{1/2}{x+1}$$

Thus

$$\int_{}^{}\frac{1}{x^2-1}dx=\int\frac{1/2}{x-1}dx-\int\frac{1/2}{x+1}$$

$$=\frac{1}{2}\ln\vert{x-1}\vert-\frac{1}{2}\ln\vert{x+1}\vert+C$$

This section teaches how to decompose

$$\frac{1}{x^{2}-1}\ into\ \frac{1/2}{x-1}-\frac{1/2}{x+1}$$

We start with a rational function \(f(x)=\frac{p(x)}{q(x)}\), where \(p\) and \(q\) do not have any common factors. We first consider the degree of \(p\) and \(q\).

- If the \(\deg(p)\geq\deg(q)\) then we use polynomial long division to divide \(q\) into \(p\) to determine a remainder \(r(x)\) where \(\deg(r)\lt\deg(q)\). We then write $$f(x)=s(x)+\frac{r(x)}{q(x)}$$ and apply partial fraction decomposition to \(\frac{r(x)}{q(x)}\)
- If the \(\deg(p)\lt\deg(q)\) we can apply partial fraction decomposition to \(\frac{p(x)}{q(x)}\) without additional work.

Partially fraction decomposition is based on an algebraic theorem that guarantees that any polynomial, including \(q\), can factor into the product of linear and irreducible quadratic factors.

An irreducible quadratic is one that cannot factor into linear terms with real coefficients. The following Key Idea states how to decompose a rational function into a sum of rational functions whose denominators are all of lower degree than \(q\).

**Key Concepts To Solve Partial Fraction Decomposition**

Let \(\frac{p(x)}{q(x)}\) be a rational function, where \(\deg(p)\lt\deg(q)\).

**Key Concept -1. Factor q(x):** Write \(q(x)\) as the product of its linear and irreducible quadratic factors of the form \((ax+b)^{m}\)

and \((ax^{2}+bx+c)^{n}\) where \(m\) and \(n\)are the highest powers of each factor that divide \(q\).

**Linear Terms:**For each linear factor of \(q(x)\) the decomposition of \(\frac{p(x)}{q(x)}\) will contain the following terms: $$\frac{A_{1}}{(ax+b)}+\frac{A_{2}}{(ax+b)^{2}}+\dots +\frac{A_{m}}{(ax+b)^{m}}$$**Irreducible Quadratic Terms:**For each irreducible quadratic factor of \(q(x)\) the decomposition of \(\frac{p(x)}{q(x)}\) will contain the following terms: $$\frac{B_{1}x+C_{1}}{(ax^{2}+bx+c)}+\frac{B_{2}x+C_{2}}{(ax^{2}+bx+c)^{2}}+\dots+\frac{B_{n}x+C_{n}}{(ax^{2}+bx+c)^{n}}$$

**Key Concept- 2. Finding The Coefficients: **\(\mathbf{A_{i}}\),\(\mathbf{B_{i}}\), and\(\mathbf{C_{i}}\):

- Set \(\frac{p(x)}{q(x)}\) equal to the sum of its linear and irreducible quadratic terms. $$\frac{p(x)}{q(x)}=\frac{A_{1}}{(ax+b)}+\dotsb\frac{A_{m}}{(ax+b)^{m}}+\frac{B_{1}x+C_{1}}{(ax^{2}+bx+c)}+\dotsb\frac{B_{n}x+C_{n}}{(ax^{2}+bx+c)^{n}}$$
- Multiply this equation by the factored form of \(q(x)\) and simplify to clear the denominators.
- Solve for the coefficients \(\mathbf{A_{i}}\),\(\mathbf{B_{i}}\), and\(\mathbf{C_{i}}\) and by
- Multiplying out the remaining terms and collecting similar powers of \(x\), equating the resulting coefficients and solving the resulting system of linear equations, or
- For \(x\) you would substitute values to eliminate terms so the simplified equation can be solved for a coefficient.

**Partial Fraction Decomposition Example**

\(\mathbf{\color{red}{Perform\ the\ partial\ fraction\ decomposition\ of\ (\frac{1}{x^{2}-1}).}}\)

The denominator can be written as the product of two linear factors: $$x^{2}-1=(x-1)(x+1)$$

Thus $$\frac{1}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsb (eq.1)$$

Using the method described in Key concept- 1(a) to solve for \(A\) and \(B\), first multiply through by \(x^{2}-1=(x-1)(x+1)\):

$$1=\frac{A(x-1)(x+1)}{x-1}+\frac{B(x-1)(x+1)}{x+1}$$

$$=A(x+1)+B(x-1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsb (eq.2)$$

$$=Ax+A+Bx-B$$

$$=(A+B)x+(A-B)\ \ \ \ \ \ \ \ \ \ \ \ collect\ like\ terms.$$

The next step is key. For clarity’s sake, rewrite the equality we have as

$$0x+1=(A+B)x+(A-B)$$

On the left, the coefficient of the term \(x\) is 0; on the right, it is \(A+B\). Since both sides are equal for all values of \(x\), we must have that \(0=A+B\). Likewise, on the left, we have a constant term of \(1\); on the right, the constant term is \(A-B\). Therefore we have \(1=A-B\).

There are two linear equations with two unknowns. It is easy to solve by hand, leading to

$${A+B=0\\A-B=1}\ \ \Longrightarrow {A=\frac{1}{2}\\B=\frac{-1}{2}}$$

Thus $$\frac{1}{x^{2}-1}=\frac{1/2}{x-1}-\frac{1/2}{x+1}$$

Note that Equations (1) and (2) are not equivalent before solving for and using the method described in Key concept-1(b). Only the second equation holds for all values of \(x\), including \(x=-1\) and \(x=1\), by continuity of polynomials. Thus, we can choose values for \(x\) that eliminate terms in the polynomial to solve for \(A\) and \(B\).

$$1=A(x+1)+B(x-1)$$

If we choose \(x=-1\),

$$1=A(0)+B(-2)$$

$$B=\frac{-1}{2}$$

Next choose \(x=1\):

$$1=A(2)+B(0)$$

$$A=\frac{1}{2}$$

This results in the same decomposition as above.

\(\mathbf{\color{red}{Use\ partial\ fraction\ decomposition\ to\ evaluate\ \int\frac{7x^{2}+31x+54}{(x+1)(x^{2}+6x+11)}dx}}\)

We begin by applying Key concept 1 since the degree of the numerator is less than the degree of the denominator. The result is:

$$\frac{7x^{2}+31x+54}{(x+1)(x^{2}+6x+11)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+6x+11}$$

Now clear the denominators.

$$7x^{2}+31x+54=A(x^{2}+6x+11)+(Bx+C)(x+1)$$

Again, we choose values of \(x\) to eliminate terms in the polynomial. If we choose \(x=-1\)

$$30=6A+(-B+C)(0),$$

$$A=5$$

Even though none of the other terms can be zeroed out, we continue by letting \(A=5\) and substituting helpful values of \(x\). Choosing \(x=0\), we notice

$$54=55+C$$

$$C=-1$$

Finally, choose \(x=1\) (any value other than \(-1\) and \(0\) can be used, \(1\) is easy to work with)

$$92=90+(B-1)(2)$$

$$B=2$$

Thus

$$\int\frac{7x^{2}+31x+54}{(x+1)(x^{2}+6x+11)}dx=\int\left(\frac{5}{x+1}+\frac{2x-1}{x^{2}+6x+11}\right)dx$$

The first term of this new integrand can be evaluated easily; it leads to \(5\ln\vert x+1\vert\). Using substitution techniques, the second term is not difficult, but it requires several steps.

The integrand \(\frac{2x-1}{x^{2}+6x+11}\) has a quadratic in the denominator and a linear term in the numerator. This leads us to try substitution.

Let \(u=x^{2}+6x+11\), so \(du=(2x+6)dx\). The numerator is \(2x-1\), not \(2x+6\), but we can get a \(2x+6\) term in the numerator by adding \(0\) in the form of \(7-7\).

$$\frac{2x-1}{x^{2}+6x+11}=\frac{2x-1+7-7}{x^{2}+6x+11}$$

$$=\frac{2x+6}{x^{2}+6x+11}-\frac{7}{x^{2}+6x+11}$$

Now we can integrate the first term with substitution, yielding \(\ln \vert x^{2}+6x+11\vert\). The final term can be integrated using arctangent. First, complete the square in the denominator:

$$\frac{7}{x^{2}+6x+11}=\frac{7}{(x+3)^{2}+2}$$

An antiderivative of the latter term can be found using Key concept- 1 and substitution:

$$\int\frac{7}{x^{2}+6x+11}dx=\frac{7}{\sqrt{2}}\tan^{-1}\left(\frac{x+3}{\sqrt{2}}\right)+C$$

Let’s start at the beginning and put all of the steps together.

$$\int{\frac{7x^{2}+31x+54}{(x+1)(x^{2}+6x+11)}}dx$$

$$=\int\left(\frac{5}{x+1}+\frac{2x-1}{x^{2}+6x+11}\right)dx$$

$$=\int{(\frac{5}{x+1})}dx+\int\frac{2x+6}{x^{2}+6x+11}dx-\int\frac{7}{x^{2}+6x+11}dx$$

$$=5\ln\vert x+1\vert+\ln\vert x^{2}+6x+11\vert-\frac{7}{\sqrt{2}}\tan^{-1}\left(\frac{x+3}{\sqrt{2}}\right)+C$$

As with many calculus problems, you should not expect to “see” the final answer immediately after seeing the problem. We break down the initial problem into smaller ones that are easier to solve. We combine the answers to the smaller problems to arrive at the final answer.

**Conclusion**

When dealing with rational functions, partial fraction decomposition is an important tool. At its heart, it is an algebraic technique, rather than a calculus one, since we are rewriting a fraction. However, it is extremely useful in the realm of calculus since it allows us to evaluate certain “complicated” integrals.

**FAQs**

**How do you decompose a partial fraction?**

The method is called “Partial Fraction Decomposition”, and it goes like this:

**Step 1:**Factor in the bottom.**Step 2:**For each of those factors, write one partial fraction.**Step 3:**Divide by the bottom so we no longer have fractions.

**What are the four cases of partial fraction decomposition?**

Special Cases of Partial Fraction Expansion

- The order of the numerator polynomial is equal to that of the denominator.
- Real roots with distinct characteristics.
- Repeated the Real Roots.
- Roots that are complex.
- In the numerator, an exponential (or other function).

**What is the purpose of partial fraction decomposition?**

Its significance lies in the fact that it provides algorithms for various computations with rational functions, including explicit computations of antiderivatives, Taylor series expansions, inverse Z-transforms, and inverse Laplace transform.

**What is the partial fraction in maths?**

Partially fractions are the fractions used to decompose rational expressions. Whenever an algebraic expression is broken down into a sum of two or more rational expressions, each part is called a partial fraction. This is, therefore, the reverse of the addition of rational expressions.

**How many fraction terms are there in the partial fraction decomposition?**

Therefore, we will get two terms in the partial fraction decomposition from this factor. The partial fraction decomposition for this expression looks like this. Remember that we just need to add the missing factors to each term.