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Last modified on July 18th, 2024

When an algebraic fraction, also known as a rational expression, is split, each fractional part with a polynomial in the denominator is called a *partial fraction*. This process of breaking such a fraction into parts is called partial fraction decomposition or partial fraction expansion.

In general, an algebraic fraction is in the form of ${f(x) = \dfrac{p(x)}{q(x)}}$, where both p(x) and q(x) are polynomials.

In the given algebraic fraction, ${\dfrac{5x+7}{x^{2}+2x-3}}$, the corresponding partial fractions are shown.

Each of the partial fractions can be summed up to obtain the same rational fraction ${\dfrac{5x+7}{x^{2}+2x-3}}$, as shown.

${\dfrac{3}{x-1}+\dfrac{2}{x+3}}$

= ${\dfrac{3\left( x+3\right) +2\left( x-1\right) }{\left( x-1\right) \left( x+3\right) }}$

= ${\dfrac{3x+9+2x-2}{x^{2}-x+3x-3}}$

= ${\dfrac{5x+7}{x^{2}+2x-3}}$

Here, ${\dfrac{3}{x-1}}$ and ${\dfrac{2}{x+3}}$ are the partial fractions of ${\dfrac{3}{x-1}+\dfrac{2}{x+3}}$

Now, let us learn how to decompose an algebraic fraction stepwise.

To decompose the partial fractions, we first factor the denominator of the given polynomial.

**1. Factoring the Denominator**

${\dfrac{5x+7}{x^{2}+2x-3}}$ = ${\dfrac{5x+7}{\left( x-1\right) \left( x+3\right) }}$

**2. Writing One Partial Fraction for Each of the Factors **

${\dfrac{5x+7}{\left( x-1\right) \left( x+3\right) }}$ = ${\dfrac{A}{x-1}+\dfrac{B}{x+3}}$

Here, A and B are the constants.

**3. Simplifying Both Sides**

${5x+7=A\left( x-1\right) +B\left( x+3\right)}$

**4. Finding the Value of Constants **

Substituting the zeros of (x – 1)(x + 3), we get

At x = 1,

5(1) + 7 = A(1 – 1) + B(1 + 3)

⇒ 12 = 4B

⇒ B = 3

At x = -3,

5(-3) + 7 = A(-3 – 1) + B(-3 + 3)

⇒ -15 + 7 = -4A

⇒ -8 = -4A

⇒ A = 2

Now, we have

${\dfrac{5x+7}{\left( x-1\right) \left( x+3\right) }}$ = ${\dfrac{2}{x-1}+\dfrac{3}{x+3}}$

This only works for proper rational expressions where the degree of the numerator is less than the degree of the denominator.

If the expression is improper where the degree of the numerator is greater than or equal to the degree of the denominator, then we perform polynomial long division first. Here, we do not factor them into complex numbers.

For example, ${\dfrac{x^{2}-2x-5}{x+2}}$ is an improper rational fraction where the degree of the numerator is 2, and the degree of the denominator is 1. To decompose this type of fraction, we follow the following steps:

**Dividing the Numerator by the Denominator **

Thus, ${\dfrac{x^{2}-2x-5}{x+2}}$ = ${x-4+\dfrac{3}{x+2}}$

Here, if the remainder produces any proper rational fractions, then we follow the usual way to decompose that fraction again.

When working with partial fractions, we sometimes encounter quadratic factors in the denominator that cannot be factored further into real numbers. These are called **irreducible quadratics**.

For example, the quadratic expression x^{2} + 1 cannot be factored into two linear factors with real coefficients because it does not have real roots (its roots are complex numbers, i and -i. So, instead of trying to factor x^{2} + 1 further, we treat it as an irreducible quadratic.

For example, (x^{2} + 9)(x^{2} – 9) can be factored into (x^{2} + 9)(x + 3)(x – 3) since the factors of (x^{2} + 9) are complex numbers. Thus, the factors could be a combination of linear factors and irreducible quadratic factors.

If we have a quadratic factor, we need to include this partial fraction:

${\dfrac{Bx+C}{quadratic}}$

Here is a list of partial fraction formulas used to decompose an algebraic fraction into its corresponding partial fractions.

Algebraic Fraction | Partial Fraction |
---|---|

1. ${\dfrac{px+q}{\left( x-a\right) \left( x-b\right) }}$ | ${\dfrac{A}{x-a}+\dfrac{B}{x-b}}$ |

2. ${\dfrac{px+q}{\left( x-a\right) ^{2}}}$ | ${\dfrac{A_{1}}{x-a}+\dfrac{A_{2}}{x-b}}$ |

3. ${\dfrac{px^{2}+qx+r}{\left( x-a\right) \left( x-b\right) \left( x-c\right) }}$ | ${\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}}$ |

4. ${\dfrac{px^{2}+qx+r}{\left( x-a\right) ^{2}\left( x-b\right) }}$ | ${\dfrac{A_{1}}{x-a}+\dfrac{A_{2}}{x-b}+\dfrac{B}{x-b}}$ |

5. ${\dfrac{px^{2}+qx+p}{\left( x-a\right) \left( x^{2}+bx+c\right) }}$ | ${\dfrac{A}{x-a}+\dfrac{Bx+C}{x^{2}+bx+c}}$ |

**Decompose the rational function ${\dfrac{x^{2}+11}{\left( x-3\right) ^{2}\left( x^{2}+1\right) }}$ into partial fractions**

Solution:

Since (x – 3)^{2} has an exponent of 2

(x^{2} + 1) is a quadratic

Then, we have

${\dfrac{x^{2}+11}{\left( x-3\right) ^{2}\left( x^{2}+1\right) }}$ = ${\dfrac{A_{1}}{x-3}+\dfrac{A_{2}}{\left( x-3\right) ^{2}}+\dfrac{Bx+C}{x^{2}+1}}$

Now, simplifying further, we get

x^{2} + 11 = A_{1}(x – 3)(x^{2} + 1) + A_{2}(x^{2} + 1) + (x – 3)^{2}(Bx + C) …..(i)

Since x = 3 is a zero of (x – 3)^{2}(x^{2} + 1),

3^{2} + 11 = A_{1}(3 – 3)(3^{2} + 1) + A_{2}(3^{2} + 1) + (3 – 3)^{2}(3B + C)

⇒ 20 = 10A_{2}

⇒ **A _{2} = 2**

Now, substituting A

x

⇒ x

⇒ x

⇒ 9 – x

Now, comparing both sides,

A

-3A

A

-3A

From (ii), A

Now, substituting A

-3B – C = 1 ⇒ 3B + C = -1 …..(vi)

-B – 6B – 6C = 0 ⇒ 7B + 6C = 0 …..(vii)

B + 3B + 3C = 3 ⇒ 4B + 3C = 3 …..(viii)

On multiplying (vi) by 7 and (vii) by 3,

21B + 7C = -7 …..(ix)

21B + 18C = 0 …..(x)

Subtracting (ix) and (x),

11C = 7

⇒

Substituting the value of C in (viii),

${4B+3\times \dfrac{7}{11}=3}$

⇒ ${4B=3-3\times \dfrac{7}{11}}$

⇒

Since A

Thus, ${\dfrac{x^{2}+11}{\left( x-3\right) ^{2}\left( x^{2}+1\right) }}$ = ${\dfrac{3}{11\left( x-3\right) }+\dfrac{2}{\left( x-3\right) ^{2}}+\dfrac{3x+7}{11\left( x^{2}+1\right) }}$

Partial fractions are widely used to integrate algebraic fractions.

Let us integrate the expression ${\dfrac{1}{\left( x-2\right) \left( x-4\right) }}$ with respect to x.

First, we will decompose the rational fraction and then perform integration. The given rational fraction can be expressed in the partial fractional form.

${\dfrac{1}{\left( x-2\right) \left( x-4\right) }}$ = ${\dfrac{A}{x-2}+\dfrac{B}{x-4}}$ …..(i)

Simplifying further, we get

1 = A(x – 2) + B(x – 4)

⇒ 1 = x(A + B) – (2A + 4B)

Comparing both sides,

A + B = 0 …..(ii)

– (2A + 4B) = 1 …..(iii)

From (i), A + B = 0 ⇒ A = -B

Substituting the value of B in (iii),

– (2A + 4B) = 1

⇒ -2(-B) – 4B = 1

⇒ -2B = 1

⇒ B = ${-\dfrac{1}{2}}$

Thus, A = ${\dfrac{1}{2}}$

Substituting these values (i),

${\dfrac{1}{\left( x-2\right) \left( x-4\right) }}$ = ${\dfrac{1}{\left( x-2\right) }-\dfrac{1}{\left( x-4\right) }}$

Now, as we know, the integral of the sum of two functions is equal to the sum of integrals of the given functions,

${\int \dfrac{dx}{\left( x-2\right) \left( x-4\right) }}$ = ${\int \dfrac{dx}{2\left( x-2\right) }-\int \dfrac{dx}{2\left( x-4\right) }}$

= ${\dfrac{1}{2}\int \dfrac{dx}{\left( x-2\right) }-\dfrac{1}{2}\int \dfrac{dx}{\left( x-4\right) }}$

= ${\dfrac{1}{2}\ln \left| x-2\right| -\dfrac{1}{2}\ln \left| x-4\right|}$

= ${\dfrac{1}{2}\ln \left| \dfrac{x-2}{x-4}\right|}$

Thus, ${\int \dfrac{dx}{\left( x-2\right) \left( x-4\right) }}$ = ${\dfrac{1}{2}\ln \left| \dfrac{x-2}{x-4}\right|}$

Last modified on July 18th, 2024