An Introduction to Partial Fractions

Do you need help wrapping your head around partial fractions? Below, we present an introduction to partial fractions and how they relate to Multivariable Calculus.

Partial fractions are a way of splitting fractions that contain polynomials into simpler fractions. Using partial fractions can help us to solve problems involving complicated fractions, including integration and differentiation.

How Do We Use Partial Fractions?

The method of splitting fractions into partial fractions is denoted as “partial fraction decomposition”. The first thing to do when decomposing partial fractions is to factor the denominator of the fraction. Next, you write out a partial fraction for each of the denominators. As you do not know what the numerator is yet, you assign a variable to each. Next, you multiply all of the terms by the common denominator. Note that this only works for proper, rational fractions. A rational function f(x) is denoted as

f(x)= \dfrac { p(x) }{ q(x) }

where p(x) and q(x) are polynomials. A rational function is proper when the degree of the denominator is greater than the degree of the numerator.

To make things clearer, let us go through some examples.

First, let’s start with something easy:

\dfrac { 10x+2 }{ { 2 }x^{ 2 }-3x+1 }

We will need to factor the polynomial in the denominator, then separate it into partial fractions:

\dfrac { 10x+2 }{ { (2x-1)(x-1) } } =\dfrac { { a }_{ 1 } }{ 2x-1 } +\dfrac { { a }_{ 2 } }{ x-1 }

Here, a₁ and a₂ are constants that need to be determined. Multiply both sides by (2x-1)(x-1) so that we no longer have fractional terms:

{ 10x+2 }={ a }_{ 1 } { ( x-1 ) } + { a }_{ 2 } { ( 2x-1 ) }

Next, we will use the roots of the polynomial (2x-1)(x-1) to find a₁ and a₂:

For x = 1, we have

{ 10(1)+2 }={ a }_{ 2 }{ (2(1)-1) }

12={ a }_{ 2 }

And for x = 0.5, we have

{ 10(0.5)+2 }={ a }_{ 1 } { (0.5-1) }

{ 7 } ={-0.5 } { a }_{ 1 }

{ -14 }={ a }_{ 1 }

This gives us

\dfrac{ 10x+2 }{ (2x-1)(x-1) } =\dfrac { -14 }{ 2x-1 } +\dfrac { 12 }{ x-1 }

If the polynomial in the denominator has a quadratic factor, then you should use the following as one of the partial fractions:

\dfrac { bx+c }{ (quadratic) }

If your fraction has a cubic polynomial as the denominator, then you would need separate it into three fractions. That is, you need a partial fraction for each exponent. For example, consider the following fraction:

\dfrac { { x }^{ 2 }+1 }{ { (x-1) }^{ 3 }x }

We can see that (x-1) appears three times in the denominator. In this case, we need to use partial fractions with denominators that have increasing powers up to three, the power of the denominator:

\dfrac { { x }^{ 2 }+1 }{ { (x-1) }^{ 3 }x } =\dfrac { a }{ x-1 } +\dfrac { b }{ { (x-1) }^{ 2 } } +\dfrac { c }{ { (x-1) }^{ 3 } } +\dfrac { d }{ x }

Next, as before, we multiply by the common denominator:

{ x }^{ 2 }+1=ax{ (x-1) }^{ 2 }+bx(x-1)+cx+{ d(x-1) }^{ 3 }

Identifying values of x to eliminate some variables, we can set x to be 0 and 1:

For x = 0,

0+1=0+0+0+{ d(0-1) }^{ 3 }

d=-1

And for x = 1,

{ 1+1 }={ 0+0+c+0 }

c=2

Now we have d and c, but we still need a and b. To find these values, we can choose two arbitrary x values and solve for a and b. We can pick small values just to make things easier:

If x = 2,

{ 4+1 }={ 2a } { (2-1) }^{ 2 }+2b(2-1)+4-{ (2-1) }^{ 3 }

5=2a+2b+3

1=a+b

And for x = -1,

1+1=-a{ (-1-1) }^{ 2 }-b(-1-1)-2-{ (-1-1) }^{ 3 }

2={ -4a }+2b+6

2={ 2a-b}

By adding our resulting equations, 1 = a + b and 2 = 2a – b, we obtain a = 1 and b = 0:

\dfrac { { x }^{ 2 }+1 }{ { (x-1) }^{ 3 }x } =\dfrac { 1 }{ x-1 } +\dfrac { 2 }{ { (x-1) }^{ 3 } } = -\dfrac { 1 }{ x }

Note that our first example relied on the roots of the polynomial to work out the unknown variables. However, sometimes using roots does not work. In the second example, we created linear equations to determine some of the unknowns. Below, we present another example, where using the roots is not possible and hence, we will gather all of the powers of x together to solve linear equations:

\dfrac { 5x }{ (x-2)({ x }^{ 2 }+x+1) }

As the denominator contains quadratics, we will use bx+c:

\dfrac { 5x }{ (x-2)({ x }^{ 2 }+x+1) } =\dfrac { a }{ (x-2) } +\dfrac { bx+c }{ ({ x }^{ 2 }+x+1) }

To solve this, we follow the same procedure as with the previous examples, and multiply both sides by the denominator of the original fraction to get

5x={ a({ x }^{ 2 }+x+1) }+{ (bx+c)(x-2) }

Then, we pick four values of x that will give us linear equations to find a, b, and c. We can start with x = 2, as this will get rid of the last term on the right:

If x = 2, then

10={ a }(4+2+1)

and

a=\dfrac { 10 }{ 7 }

There is no easy choice for x to target the first term on the right-hand side, so we will equate the coefficients instead:

5x={ a({ x }^{ 2 }+x+1) }+{ (bx+c)(x-2) }

=a{ x }^{ 2 }+ax+a+b{ x }^{ 2 }-2bx+cx-2c

={ { x }^{ 2 }(a+b) }+{ x(a-2b+c) }+(a-2c)

As there is no term with x² on the left-hand side, we can say that

{ a+b }=0

We know that

a=\dfrac { 10 }{ 7 }

Hence,

b=-{ \dfrac { 10 }{ 7 } }

Similarly, since the left-hand side has no constants, we can say that

{ a-2c }=0

which means that

c=\dfrac { 5 }{ 7 }

Putting this all together, we have

\dfrac { 5x }{ (x-2)({ x }^{ 2 }+x+1) } =\dfrac { \dfrac { 10 }{ 7 } }{ (x-2) } +\dfrac { \dfrac { -10x }{ 7 } +\dfrac { 5 }{ 7 } }{ ({ x }^{ 2 }+x+1) }

=\dfrac { 10 }{ 7(x-2) } +\dfrac { 5(-2x+1) }{ 7({ x }^{ 2 }+x+1) }

As we mentioned above, using partial fractions to simplify a function only works with proper, rational functions. When dealing with improper rational functions, i.e. when the degree of the numerator is greater than the degree of the denominator, we first need to perform long division on the polynomial in the numerator of the fraction. For example, let’s say we that we want to perform partial fraction decomposition on the following improper rational function:

\dfrac { p(x) }{ q(x) } =\dfrac { { x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 }{ { x }^{ 3 }+{ 3x }^{ 2 }+7x+5 }

As you can see, the degree of the numerator is 4 and the degree of the denominator is 3. Consequently, we shall perform long division on p(x), such that

\dfrac { p(x) }{ q(x) } =s(x)+\dfrac { r(x) }{ q(x) }

where s(x) and r(x) are polynomials, and the degree of r(x) is less than the degree of q(x).

So, we start with

\begin{array}{c} x { x }^{ 3 }+3{ x }^{ 2 }+7x+5 \overline{ ){ x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 } \end{array}

We divide first by x because multiplying the denominator by x will give the denominator the degree of the numerator. Next, we subtract the product of x and x³ + 3x² + 7x + 5:

\begin{array}{r} x { x }^{ 3 }+3{ x }^{ 2 }+7x+5 \overline{ ){ x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 } - ({ x }^{ 4 }+3{ x }^{ 3 }+7{ x }^{ 2 }+5x)= 0+2{ x }^{ 3 }+9{ x }^{ 2 }+21x+22 \end{array}

Hence, we have the remainder 2x³ + 9x² + 21x + 22. We need to multiply the denominator by 2 to get to the highest degree in the remainder.

\begin{array}{r} x+2 { x }^{ 3 }+3{ x }^{ 2 }+7x+5 \overline{ ){ x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 }-({ x }^{ 4 }+3{ x }^{ 3 }+7{ x }^{ 2 }+5x)=0+2{ x }^{ 3 }+9{ x }^{ 2 }+21x+22 \end{array}

Next, as in the previous step, we subtract the product of 2 and the denominator:

\begin{array}{r} x+2 { x }^{ 3 }+3{ x }^{ 2 }+7x+5 \overline{ ){ x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 }-({ x }^{ 4 }+3{ x }^{ 3 }+7{ x }^{ 2 }+5x)=0+2{ x }^{ 3 }+9{ x }^{ 2 }+21x + 22 -(2{ x }^{ 3 }+6{ x }^{ 2 }+14x+10) =0+3{ x }^{ 2 }+7x+12 \end{array}

Now, we have a remainder of 3x² + 7x + 12, which has a degree of 2; this is less than the degree of the denominator, so we can stop the long division here. In summary, we subtracted the numerator by

x({ x }^{ 3 }+3{ x }^{ 2 }+7x+5)

Followed by

2(x^{ 3 }+3{ x }^{ 2 }+7x+5)

i.e.

{ x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22-x({ x }^{ 3 }+3{ x }^{ 2 }+7x+5)-2({ x }^{ 3 }+3{ x }^{ 2 }+7x+5)=3{ x }^{ 2 }+7x+12

With some rearranging, we have

{ x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22-{ (x+2)({ x }^{ 3 }+3{ x }^{ 2 }+7x+5) }=3{ x }^{ 2 }+7x+12

And with even more rearranging, we get the original improper rational fraction on the left-hand side:

\dfrac { { x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 }{ { x }^{ 3 }+3{ x }^{ 2 }+7x+5 } =x+2+\dfrac { 3{ x }^{ 2 }+7x+12 }{ { x }^{ 3 }+3{ x }^{ 2 }+7x+5 }

The right-hand side of the above equation fits the form

s(x)+\dfrac { r(x) }{ q(x) }

As the numerator of the fraction on the right-hand side is of a smaller degree compared to the denominator, we have a proper, rational function. Note that q(x) equals our remainder in the long division, and s(x) is the final quotient. Now that we have our proper, rational function, we are ready to factor the denominator, x³ + 3x² + 7x + 5. To do this, let’s try to figure out the roots of the denominator. Any integer root must be a factor of 5, so let’s first try 1:

{ 1 }^{ 3 }+3 \times { 1 }^{ 2 }+7 \times 1+5=16

As the result is not zero, 1 is not a root. Now, we can try -1:

{ (-1 })^{ 3 }+3 \times { (-1 })^{ 2 }+7 \times (-1)+5=0

Hence, -1 is a root; therefore, (x+1) is a factor, and we can use long division to factor this out of the denominator.

\begin{array}{r} { x }^{ 2 }+2x+5 x+1 \overline{ ){ x }^{ 3 }+3{ x }^{ 2 }+7x+5 }-({ x }^{ 3 }+{ x }^{ 2 })=0+2{ x }^{ 2 }+7x+5 -(2{ x }^{ 2 }+2x) =5x+5-(5x+5)=0 \end{array}

Here, x²(x+1) was subtracted, as x + 1 should be multiplied by x² to reach the same degree as x³+3x²+7x+5.  For the same reason, 2x(x+1) and 5(x+1) were subtracted, leaving us with 0. Writing this out explicitly:

{ x }^{ 3 }+3{ x }^{ 2 }+7x+5-{ x }^{ 2 }(x+1)-2x(x+1)-5(x+1)=0

And with a bit of rearranging, we end up with:

{ x }^{ 3 }+3{ x }^{ 2 }+7x+5=({ x }^{ 2 }+2x+5)(x+1)

Now, all we need to do is find the roots of the quadratic factor on the right-hand side. However, it turns out that x²+2x+5 has no real roots (HINT: to validate this, use the quadratic formula to try and find the roots. You will see that you end up with the square root of a negative number). Hence, we cannot factor this quadratic and shall keep it as it is. If we put everything together, we have

\dfrac { { x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 }{ { x }^{ 3 }+3{ x }^{ 2 }+7x+5 } =x+2+ \dfrac { 3{ x }^{ 2 }+7x+12 }{ ({ x }^{ 2 }+2x+5)(x+1) } =x+2+ \dfrac { a }{ x+1 } + \dfrac { bx+c }{ { x }^{ 2 }+2x+5 }

To find a, b, and c, we multiply both sides by the denominator, as with previous examples:

\dfrac { a }{ x+1 } + \dfrac { bx+c }{ { x }^{ 2 }+2x+5 } = \dfrac { a({ x }^{ 2 }+2x+5)+(bx+c)(x+1) }{ (x+1)({ x }^{ 2 }+2x+5) } = \dfrac { 3{ x }^{ 2 }+7x+12 }{ ({ x }^{ 2 }+2x+5)(x+1) }

Now we can equate the numerators of the middle and last fraction given above:

a({ x }^{ 2 }+2x+5)+(bx+c)(x+1)=3{ x }^{ 2 }+7x+12

Rearranging and collected the terms together:

{ x }^{ 2 }(a+b)+x(2a+b+c)+(5a+c)=3{ x }^{ 2 }+7x+12

We have:

(a+b)=3

(2a+b+c)=7

(5a+c)=12

Using substitution, we can solve the three equations above and determine that a = 2, b = 1, and c = 2. Finally, we can substitute the values of these constants into the partial fraction decomposition from above:

\dfrac { { x }^{ 4 }+5{ x }^{ 3 }+16{ x }^{ 2 }+26x+22 }{ { x }^{ 3 }+3{ x }^{ 2 }+7x+5 } = x+2+ \dfrac { 2 }{ x+1 } + \dfrac { x+2 }{ { x }^{ 2 }+2x+5 }

Partial Fractions and Multivariable Calculus

Now that we have gone through the basics of partial fractions, we can go on to see how they can be used in multivariable calculus. In general, partial fractions can be used when dealing with multivariable functions for integration, differentiation, series expansion, differential equations, etc. when complex fractions must be simplified to solve the problem.

For something more specific, partial fractions are used when dealing with the inverse Laplace transform. The inverse Laplace transform, coupled with the Laplace transform, is used to rewrite and solve difficult differential equations. The Laplace transform is applied to complicated ordinary different equations or partial differential equations to simplify them into simple algebraic problems. The inverse Laplace transform is then applied to derive the solution of the original, more complicated differential equations.

The Laplace transform can be defined as follows:

F(s)=\int _{ 0 }^{ \infty }{ f(t){ e }^{ -st }dt }

Here, f(t) is a real function and F(s) = L(f(t)), where L is the Laplace transform operator. Thus, L^-1 is the inverse Laplace transform, with f(t)= L^-1(F(s)). Partial fractions can help invert complicated fractions for the inverse Laplace transform. Usually, F(s) is a complicated fraction that needs to be simplified before its inverse can be computed, just as in the examples above.

After reading this review article, we hope that you are now able to understand how to solve all kinds of partial fractions, and have an idea of how they are related to multivariable calculus. We have given you a wide spectrum of examples to work through so that you are prepared to face all types of partial fraction problems.

Let’s put everything into practice. Try this Multivariable Calculus practice question:

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