Integration Using Partial Fractions

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Partial fraction decomposition allows us to undo the process of adding fractions with polynomial denominators together. We can also apply this to integration. You might be asked to integrate an expression that looks tricky, and you should check if you can factorise the denominator. This means that we can split one big fraction into multiple smaller fractions, and from here, the integral will appear more straightforward.

When we do this, we should be left with multiple integrals that are easy to integrate. Usually, these are left as a linear factor. However, we may have to use a different method when we are left with an irreducible polynomial that doesn't factorise.

Recap of partial fraction decomposition

Integrating using partial fractions is used for expressions in the form of a fraction. Before we begin, we define the degree of a polynomial to be the order of the highest order term, i.e. the degree of \( x^4 + 3x +1\) is \(4\), and the degree of \(x + x^8 - 5\) is \(8\). The first thing we need to check is whether the degree of the numerator is less than the degree of the denominator. If so, we can continue.

If the degree of the numerator is more than the degree of the denominator, we must first perform algebraic long division.

The next step is to factorise the denominator. First, check for linear factors. This involves finding roots (via the factor theorem - if \(f (x)\) is a polynomial of degree \(n \ge 1\) and \(a\) is any real number, then \(x-a\) is a factor of \(f(x)\), and then this will give the linear factors. If the degree of the denominator is greater than or equal to four, then we must also look for quadratic factors.

\(x^4 + 5x^2 + 4\) has no real roots, but

\[(x^2 + 1)(x^2 + 4) = x^4 + 5x^2 + 4\]

so we must check this

Note: if the degree were greater than or equal to six, then we would have to check for cubic factors; however, these examples are few and far between, so let's focus on the examples where the degree of the denominator is less than or equal to \(5\).

We now need to note if there are any repeat factors. This affects how we write the decomposed fraction. If a factor doesn't repeat, we only need to consider the factor as a separate numerator. If it appears more than once, we need to consider all its possible multiples. The degree of the polynomial on the numerator should always be one less than the polynomial in the denominator unless it is repeated, where it is the same as if it were an unrepeated root.

We can then write the form of our decomposed fraction.

Suppose we were to look for the partial fraction decomposition of

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

This means we would be looking for an expression of the form

\[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+1}.\]

Once we have the form of the decomposed fraction, we can now solve for each unknown coefficient. We multiply through by the denominator and then equate equivalent coefficients.

Find the partial fraction decomposition of

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

From above, we know the form of this should be

\[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+1}.\]

Let's multiply through by \( x^3 (x^2 + 1) \) to get

\[ 1 = Ax^2(x^2+1) + Bx(x^2+1) + C(x^2+1) + (Dx+E)x^3,\]

which then simplifies to

\[1 = (A+D)x^4 + (B+E)x^3 + (A+C)x^2 + Bx + C.\]

Comparing coefficients, we get

\[ \begin{align} & A + D = 0 \\ & B + E = 0 \\ & A + C = 0 \\ & B = 0 \\ &C = 1. \end{align}\]

Solving this (we have two values trivially, and then filling these in), we get \(B = E = 0\), \(D = C = 1\), and \(A = -1\). This then gives us

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

Using partial fractions in integrals

We have recapped using partial fractions, which will help us evaluate integrals. See below:

Integrate

\[ \int \frac{1}{x^2+1} \, \mathrm{d}x.\]

Our first step here is to perform partial fraction decomposition. By the difference of two squares, we know that

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

This means that the expected form of the partial fraction will be

\[ \frac{A}{x-1} + \frac{B}{x+1}.\]

Now let us equate the two sides, which leaves us with

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

Multiplying through by \(x^2-1\) we get

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

This implies that \(A + B = 0\) and \(A - B = 1\). This gives \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\).

You can now apply this to the integral, which gives

\[ \int \frac{1}{x^2+1} \, \mathrm{d}x = \frac{1}{2} \int \frac{1}{x-1} \, \mathrm{d}x - \frac{1}{2} \int \frac{1}{x+1} \, \mathrm{d}x. \]

This integral is now easy to integrate. This evaluates to

\[ \frac{1}{2} \left( \ln |x-1| - \ln |x+ 1|\right) + C = \frac{1}{2} \ln\left| \frac{x-1}{x+1} \right| + C.\]

Integrate

\[ \int \frac{1}{x^3(x^2+1)} \, \mathrm{d}x.\]

From the above examples, we can reduce this to

\[ -\int \left(\frac{1}{x} +\frac{1}{x^3} + \frac{x}{x^2+1}\right) \, \mathrm{d}x. \]

Define

\[ \begin{align} I &= \int \frac{1}{x} \, \mathrm{d}x , \\ J &= \int \frac{1}{x^3}\, \mathrm{d}x , \\ K &= \int \frac{x}{x^2+1} \, \mathrm{d}x .\end{align}\]

\(I\) is a standard integral, evaluating to \( \ln |x| + C_1\). \(J\) can be evaluated using the formula for integrating a polynomial and is given as

\[ -\frac{1}{2}x^{-2} + C_2.\]

\(K\) can be evaluated using a substitution. Let \(u = x^2 + 1\), then

\[ \mathrm{d}x = \frac{1}{2x} \mathrm{d}u ,\]

which gives

\[ \begin{align} K &= \int \frac{x}{x^2+1} \, \mathrm{d}x \\ &= \frac{1}{2} \int\frac{1}{u} \, \mathrm{d}u \\ &= \frac{1}{2}\ln |x^2+1| + C_3. \end{align}\]

We can now combine these, as

\[ \begin{align} \int \frac{1}{x^3(x^2+1)} \, \mathrm{d}x &= I + J + K \\ & = \ln |x| -\frac{1}{2}x^{-2} + \frac{1}{2}\ln |x^2+1| + C. \end{align} \]