Partial Fraction Decomposition

Practice

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Let's say we're trying to compute the following integral

$$\int_{}^{} \frac{2}{x^{2}-5x+6} ,dx $$

Since the polynomial is in the denominator of a fraction, we will have to rework this problem algebraically before taking the antiderivative. Partial Fraction Decomposition is this method we will use to create a sum of easier integrals.

Step 1

$$\int_{}^{} \frac{2}{\color{red}x^{2} - 5x + 6},\color{black}dx = \int_{}{}\frac{2}{\color{red}(x-2)(x-3)},dx$$

Factor the denominator of the rational expression (if necessary)

Step 2

$$ \color{red}\frac{2}{(x-2)(x-3)} \color{black}= \color{blue}\frac{A}{(x-2)} + \frac{B}{(x-3)}$$

If this is your first time working with Partial Fraction Decomposition, this will be confusing. We've made up two new variables, A, and B. This step states that we can rewrite a single rational expression with two factors multiplied together in the denominator as two separate rational expressions that have uknown constants as numerators. We will go into more detail about this step later in the article.

Step 3

$$ \color{blue}(x-2)(x-3)\color{red}*\color{black}\frac{2}{\color{blue}(x-2)(x-3)\color{black}} = (\frac{A}{(x-2)} + \frac{B}{(x-3)})\color{red}*\color{blue}(x-2)(x-3)$$

> Multiply both sides by the denominator on the left side of the equation, this will completely cancel all the denominators in the equation.

$$ \require{cancel}\cancel{(x-2)}\cancel{(x-3)}*\frac{2}{\cancel{(x-2)}\cancel{(x-3)}} = (\frac{A}{\cancel{(x-2)}} + \frac{B}{\cancel{(x-3)}})*(x-2)(x-3) $$

At this point, you should have a relatively simple equation

$$ 2 = A(x-3) + B(x-2) $$

Step 4

This equation needs to be true for all values of x, including x = 2 and x = 3, We can substitute those values in for x to find A and B

$$ 2 = A(x-3) + B(x-2) $$ $$ \begin{array}{c|c} x = 2 & x = 3 \\ 2 = A((2)-3) + B((2)-2) & 2 = A((3)-3) + B((3)-2) \\ 2 = A(-1) & 2 = B(1) \\ A = -2 & B = 2 \end{array} $$

Step 5

$$ \frac{2}{x^{2} - 5x + 6} = \frac{-2}{x-2} + \frac{2}{x-3} $$

We've now shown algebraically that the expression in the original problem can be described as a sum of two simpler rational expressions, we end up with the following

$$ \int (\frac{-2}{x-2} + \frac{2}{x-3}) dx $$

Step 6

Now we just need to integrate!

$$ \displaylines{\int \frac{-2}{x-2} + \frac{2}{x-3} dx = \int \frac{-2}{x-2}dx + \int \frac{2}{x-3}dx \\ \\ = -2ln(x-2) + 2ln(x-3) + C} $$

We're done! but there's more to discuss, particularly Step 2, which we hand-waved over earlier

The Algebra at Work

Many times taking difficult integrals is just a matter of re-writing the original problem in a different form using nothing but algebra. The integration itself was one of the easier steps. The algebra however will get more complicated. Let's revisit Step 2 from the introduction problem.

$$ \frac{2}{(x-2)(x-3)} = \frac{A}{(x-2)} + \frac{B}{(x-3)} $$

Why were we allowed to do this? Let's work backwards and try to recombine the two rational expressions using common denominators

$$ \displaylines{ \frac{2}{(x-2)(x-3)} = \frac{A}{(x-2)} + \frac{B}{(x-3)} = \frac{-2}{x-2} + \frac{2}{x-3} \\ \
\frac{2}{(x-2)(x-3)} = \frac{-2}{x-2}*\frac{x-3}{x-3} + \frac{2}{x-3}*\frac{x-2}{x-2} \\ \
\frac{2}{(x-2)(x-3)} = \frac{-2(x-3)}{(x-2)(x-3)} + \frac{2(x-2)}{(x-2)(x-3)} \\ \
\frac{2}{(x-2)(x-3)} = \frac{-2(x-3) + 2(x-2)}{(x-2)(x-3)} \\ \
\frac{2}{(x-2)(x-3)} = \frac{-2x + 6 + 2x - 4}{(x-2)(x-3)} \\ \
\frac{2}{(x-2)(x-3)} = \frac{2}{(x-2)(x-3)} }$$

Essentially, Partial Fraction Decomposition is just the process of combining fractions using common denominators in reverse. This will get more complicated as we introduce polynomials of higher degree in the numerator and denominator of the function. For example, if we have a repeated factor, such as the following

$$ \frac{2}{(x+1)^{2}} $$

We would use a different form of deconstruction, in this case

$$ \frac{2}{(x+1)^{2}} = \frac{A}{(x+1)} + \frac{B}{(x+1)^{2}} $$