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## Section1.10Partial Fractions

Partial fractions is the name given to a technique of integration that may be used to integrate any rational function  1 Recall that a rational function is the ratio of two polynomials.. We already know how to integrate some simple rational functions

\begin{align*} \int \frac{1}{x}\dee{x} &= \log|x|+C & \int \frac{1}{1+x^2} \dee{x} &= \arctan(x) +C \end{align*}

Combining these with the substitution rule, we can integrate similar but more complicated rational functions:

\begin{align*} \int \frac{1}{2x+3}\dee{x} &= \frac{1}{2} \log|2x+3| +C & \int \frac{1}{3+4x^2}\dee{x} &= \frac{1}{2\sqrt{3}} \arctan\left( \frac{2x}{\sqrt{3}} \right) +C \end{align*}

By summing such terms together we can integrate yet more complicated forms

\begin{align*} \int \left[ x + \frac{1}{x+1} + \frac{1}{x-1} \right]\dee{x} &= \frac{x^2}{2} + \log|x+1| + \log|x-1| +C \end{align*}

However we are not (typically) presented with a rational function nicely decomposed into neat little pieces. It is far more likely that the rational function will be written as the ratio of two polynomials. For example:

\begin{gather*} \int \frac{x^3+x}{x^2-1}\dee{x} \end{gather*}

In this specific example it is not hard to confirm that

\begin{align*} x+\frac{1}{x+1} +\frac{1}{x-1} &=\frac{x(x+1)(x-1) +(x-1) +(x+1)}{(x+1)(x-1)} =\frac{x^3+x}{x^2-1} \end{align*}

and hence

\begin{align*} \int \frac{x^3+x}{x^2-1}\dee{x} &= \int \left[ x + \frac{1}{x+1} + \frac{1}{x-1} \right]\dee{x} \\ &= \frac{x^2}{2} + \log|x+1| + \log|x-1| +C \end{align*}

Of course going in this direction (from a sum of terms to a single rational function) is straightforward. To be useful we need to understand how to do this in reverse: decompose a given rational function into a sum of simpler pieces that we can integrate.

Suppose that $N(x)$ and $D(x)$ are polynomials. The basic strategy is to write $\frac{N(x)}{D(x)}$ as a sum of very simple, easy to integrate rational functions, namely

1. polynomials — we shall see below that these are needed when the degree  2 The degree of a polynomial is the largest power of $x\text{.}$ For example, the degree of $2x^3+4x^2+6x+8$ is three. of $N(x)$ is equal to or strictly bigger than the degree of $D(x)\text{,}$ and
2. rational functions of the particularly simple form $\frac{A}{(ax+b)^n}$ and
3. rational functions of the form $\frac{Ax+B}{(ax^2+bx+c)^m}\text{.}$

We already know how to integrate the first two forms, and we'll see how to integrate the third form in the near future.

To begin to explore this method of decomposition, let us go back to the example we just saw

\begin{align*} x+\frac{1}{x+1} +\frac{1}{x-1} &=\frac{x(x+1)(x-1) +(x-1) +(x+1)}{(x+1)(x-1)} =\frac{x^3+x}{x^2-1} \end{align*}

The technique that we will use is based on two observations:

1. The denominators on the left-hand side of are the factors of the denominator $x^2-1=(x-1)(x+1)$ on the right-hand side.
2. Use $P(x)$ to denote the polynomial on the left hand side, and then use $N(x)$ and $D(x)$ to denote the numerator and denominator of the right hand side. That is

\begin{align*} P(x)&=x & N(x)&= x^3+x & D(x)&= x^2-1. \end{align*}

Then the degree of $N(x)$ is the sum of the degrees of $P(x)$ and $D(x)\text{.}$ This is because the highest degree term in $N(x)$ is $x^3\text{,}$ which comes from multiplying $P(x)$ by $D(x)\text{,}$ as we see in

\begin{align*} x + \frac{1}{x+1} + \frac{1}{x-1} &=\frac{ \overbrace{x}^{P(x)} \overbrace{(x+1)(x-1)}^{D(x)} + (x-1) + (x+1) } {(x+1)(x-1)} =\frac{x^3+x}{x^2-1} \end{align*}

More generally, the presence of a polynomial on the left hand side is signalled on the right hand side by the fact that the degree of the numerator is at least as large as the degree of the denominator.