Partial Fractions

Partial fraction decomposition is a technique used to write a rational function as the sum of simpler rational expressions.

\[\frac{2}{x^2-1} \Rightarrow \frac{1}{x-1} - \frac{1}{x+1}.\]

Partial fraction decomposition is a useful technique for some integration problems involving rational expressions. Partial fraction decomposition is also useful for evaluating telescoping sums. It is the basis for a proof of Euler's formula by finding the antiderivative of a rational expression in two different ways.

Main Article: Partial Fractions - Linear Factors

A partial fraction decomposition has linear factors when the denominator can be factored into linear polynomials, each of which has multiplicity 1.

Find the partial fraction decomposition of the following rational expression:

\[\frac{2x+1}{x^2-x-6}.\]

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The denominator can be factored:

\[x^2-x-6 = (x+2)(x-3).\]

Then the form of the partial fraction decomposition is

\[\frac{2x+1}{x^2-x-6} = \frac{A}{x+2}+\frac{B}{x-3}.\]

Solving for the coefficients gives

\[\begin{align} \frac{2x+1}{x^2-x-6} &= \frac{A(x-3)+B(x+2)}{(x+2)(x-3)}\\\\\\ 2x+1 &= A(x-3)+B(x+2) \\ &= (A+B)x+(2B-3A) \\\\ A+B &= 2 \\ 2B-3A &= 1. \end{align}\]

Solving this system of equations gives \(A=\frac{3}{5}\) and \(B=\frac{7}{5}.\) Then the partial fraction decomposition is

\[\frac{2x+1}{x^2-x-6}=\frac{1}{5}\left(\frac{3}{x+2}+\frac{7}{x-3}\right).\ _\square\]

Main Article: Partial Fractions - Cover Up Rule

The cover-up rule is a technique to efficiently compute the coefficients of a partial fraction decomposition with linear factors.

Find the partial fraction decomposition of the following rational expression:

\[\frac{2x}{x^2-9x+18}.\]

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The denominator can be factored:

\[x^2-9x+18=(x-3)(x-6).\]

Then the partial fraction decomposition form is

\[\frac{2x}{(x-3)(x-6)}=\frac{A}{x-3}+\frac{B}{x-6}.\]

To solve for \(A,\) "cover up" the \(x-3\) factor and substitute \(x=3\) into the original expression:

\[A = \frac{2x}{x-6} = \frac{2\cdot 3}{3-6} = -2.\]

To solve for \(B,\) "cover up" the \(x-6\) factor and substitute \(x=6\) into the original expression:

\[B = \frac{2x}{x-3} = \frac{2\cdot 6}{6-3} = 4.\]

Then the partial fraction decomposition is

\[\frac{2x}{x^2-9x+18}=\frac{4}{x-6}-\frac{2}{x-3}.\ _\square\]

Main Article: Partial Fractions - Repeated Factors

A partial fraction has repeated factors when one of the factors has multiplicity greater than 1, in other words, when one of the factors is raised to a power 2 or greater.

Find the partial fraction decomposition of the following rational expression:

\[\frac{x-4}{x^2-10x+25}.\]

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The denominator factors as a perfect square:

\[x^2-10x+25=(x-5)^2.\]

Then the partial fraction decomposition form is

\[\frac{x-4}{x^2-10x+25}=\frac{A}{x-5}+\frac{B}{(x-5)^2}.\]

Solve for the coefficients:

\[\begin{align} \frac{x-4}{x^2-10x+25} &= \frac{A(x-5)+B}{(x-5)^2} \\\\\\ x-4 &= A(x-5)+B \\ x-4 &= Ax +(B-5A)\\\\ 1 &= A \\ -4 &= B-5A. \end{align}\]

This gives \(A=1\) and \(B=1\) as the coefficients. Then the partial fraction decomposition is

\[\frac{x-4}{x^2-10x+25} = \frac{1}{x-5}+\frac{1}{(x-5)^2}.\ _\square\]

Main Article: Partial Fractions - Irreducible Quadratics

A partial fraction decomposition has irreducible quadratics when one of the factors is a quadratic that does not have rational roots.

Find the partial fraction decomposition of the following rational expression:

\[\frac{1}{x^3-1}.\]

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The denominator can be factored as a difference of cubes:

\[x^3-1=(x-1)(x^2+x+1).\]

The quadratic has complex roots, so it cannot be factored any further. Then the partial fraction decomposition form is

\[\frac{1}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}.\]

Solving for the coefficients gives

\[\begin{align} \frac{1}{x^3-1} &= \frac{A(x^2+x+1)+(Bx+C)(x-1)}{(x-1)(x^2+x+1)}\\\\\\ 1 &= A(x^2+x+1)+(Bx+C)(x-1) \\ 1 &= Ax^2+Ax+A+Bx^2+(C-B)x-C \\ 1 &= (A+B)x^2+(A+C-B)x+(A-C) \\\\ A+B &= 0 \\ A+C-B &= 0 \\ A-C &= 1. \end{align}\]

Solving this system of equations gives \(A=\frac{1}{3},\) \(B=-\frac{1}{3},\) and \(C=-\frac{2}{3}.\) Then partial fraction decomposition is

\[\frac{1}{x^3-1}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{x+2}{x^2+x+1}\right).\ _\square\]

Main Article: Integration with Partial Fractions

Partial fraction decomposition is often used to find integrals of rational functions. It is useful when the \(u\)-substitution technique does not work.

Find the indefinite integral

\[\int{\frac{dx}{x^2+3x+2}}.\]

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First note that there is no obvious \(u\)-substitution that can be done to simplify the integral. Instead, use partial fraction decomposition to write the expression as the sum of two rational expressions:

\[\frac{1}{x^2+3x+2}=\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}.\]

Then the integral becomes

\[\begin{align} \int{\frac{dx}{x^2+3x+2}} &= \int{\frac{dx}{x+1}}-\int{\frac{dx}{x+2}} \\ \\ &=\ln|x+1|-\ln|x+2|+C, \end{align}\]

where \(C\) is the constant of integration. \(_\square\)

Main Article: Telescoping Series - Sum

A series of rational expressions can sometimes contain a hidden telescoping sum. Using partial fraction decomposition can often reveal this telescoping sum so that evaluating the sum becomes much easier.

Evaluate the following sum:

\[\sum\limits_{k=1}^{40}{\frac{2}{k^2+4k+3}}.\]

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At first glance, it's difficult to discern any pattern in the terms of the series:

\[\sum\limits_{k=1}^{40}{\frac{2}{k^2+4k+3}}=\frac{1}{4}+\frac{2}{15}+\frac{1}{12}+\frac{2}{35}+\cdots+\frac{1}{840}+\frac{2}{1763}.\]

Performing partial fraction decomposition on the rational expression gives

\[\frac{2}{k^2+4k+3}=\frac{1}{k+1}-\frac{1}{k+3}.\]

Then the sum becomes

\[\sum\limits_{k=1}^{40}\left(\frac{1}{k+1}-\frac{1}{k+3}\right).\]

From writing out the terms of the sum, a pattern emerges:

\[\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\cdots+\frac{1}{39}-\frac{1}{41}+\frac{1}{40}-\frac{1}{42}+\frac{1}{41}-\frac{1}{43}.\]

It's clear that many of the terms will cancel out. The exact value of the series can be found by re-writing the sums:

\[\begin{array}{ccccc} \sum\limits_{k=1}^{40}\left(\frac{1}{k+1}-\frac{1}{k+3}\right) & = & \sum\limits_{k=2}^{41}{\frac{1}{k}} & - & \sum\limits_{k=4}^{43}{\frac{1}{k}} \\ \\ & = & \frac{1}{2}+\frac{1}{3}+\sum\limits_{k=4}^{41}{\frac{1}{k}} & - & \sum\limits_{k=4}^{41}{\frac{1}{k}}-\frac{1}{42}-\frac{1}{43} \\ \\ & = & \frac{1}{2}+\frac{1}{3} & - & \frac{1}{42}-\frac{1}{43} \\ \\ & = & \frac{710}{903}.\ _\square \end{array}\]

Partial fraction decomposition is the basis for a proof of Euler's formula.

Euler's Formula:

\[e^{i\theta}=\cos{\theta}+i\sin{\theta}.\ _\square\]

The following proof uses integration, trigonometric identities, and logarithmic identities.

Consider the rational expression

\[\frac{1}{1+x^2}.\]

Now consider the antiderivative of this expression. One possible method to obtain the antiderivative of this expression is to apply trigonometric substitution. This method yields

\[\int{\frac{\mathrm{d} x}{1+x^2}}=\arctan{x} + \text{C}_{1}, \qquad (1)\]

where \(\text{C}_{1}\) is the constant of integration.

Another possible method to obtain the antiderivative of this expression is to apply partial fraction decomposition. The denominator of the expression does not have real roots, but it can be factored into complex roots:

\[1+x^2=(1+ix)(1-ix).\]

Applying partial fraction decomposition gives the equivalent expression

\[\frac{1}{1+x^2}=\frac{1}{2}\left(\frac{1}{1+ix}+\frac{1}{1-ix}\right).\]

Thus, an alternative antiderivative of the expression is

\[\begin{align} \int{\frac{\mathrm{d}x}{1+x^2}} &= \frac{1}{2}\int\left(\frac{1}{1+ix}+\frac{1}{1-ix}\right) \mathrm{d}x \\ \\ &= \frac{1}{2i}\big[\ln(1+ix)-\ln(1-ix)\big] + \mathrm{C}_{2} \\ \\ &= \frac{1}{2i}\left[\ln\left(\frac{1+ix}{1-ix}\right)\right] + \mathrm{C}_{2}, \qquad (2) \end{align}\]

where \(\text{C}_{2}\) is the constant of integration.

Because these antiderivatives are of the same expression, they are equivalent, i.e. \((1)=(2):\)

\[\arctan{x}=\frac{1}{2i}\left[\ln\left(\frac{1+ix}{1-ix}\right)\right] + \text{C} .\]

where \( \text{C} = \text{C}_{2} - \text{C}_{1} \).

Putting \( x=0 \), we find that \(\text{C} = 0\), which implies

\[\arctan{x}=\frac{1}{2i}\left[\ln\left(\frac{1+ix}{1-ix}\right)\right]. \]

Let \(x=\tan{\theta}\), then substituting gives

\[\theta=\frac{1}{2i}\left[\ln\left(\frac{1+i\tan{\theta}}{1-i\tan{\theta}}\right)\right].\]

Further simplifying and use of identities gives

\[\begin{align} i\theta &= \frac{1}{2}\left[\ln\left(\frac{(1+i\tan{\theta})^2}{1+\tan^2{\theta}}\right)\right] \\ \\ &= \frac{1}{2}\left[\ln\left(\frac{(1+i\tan{\theta})^2}{\sec^2{\theta}}\right)\right] \\ \\ &= \ln\left(\frac{1+i\tan{\theta}}{\sec{\theta}}\right) \\ \\ &= \ln\left(\cos{\theta} + i\sin{\theta}\right). \end{align}\]

Then, writing the equation in equivalent exponential form gives Euler's formula:

\[e^{i\theta}=\cos{\theta}+i\sin{\theta}.\ _\square\]