Theory

Theory 1

A rational function is a ratio of polynomials, for example:

$$\frac{P(x)}{Q(x)}=\frac{5x^2+x-28}{x^3-4x^2+x+6}$$

Partial fraction decomposition

The partial fraction decomposition of a rational function is a way of writing it as a sum of simple terms, like this:

$$\frac{3x^3-5x^2-6x+20}{x^4-3x^3+4x}=-\frac{2}{x+1}+\frac{2}{{(x-2)}^2}+\frac{5}{x}$$

Allowed denominators:

• Linear, e.g. $x-a$, or linear power, e.g. ${(x-a)}^n$
• Quadratic, e.g. $x^2+bx+c$, or quadratic power, e.g. ${(x^2+bx+c)}^n$
  • Condition: quadratics must be irreducible. (No roots, i.e. $b^2<4c$.)

Allowed numerators: constant (over linear power) or linear (over quadratic power)

These are allowed as simple terms in partial fraction decompositions:

$$\frac{1}{x^2+1},\frac{2x+1}{x^2+5},\frac{7}{5x-8},\frac{1}{x},\frac{1}{x^3}$$

These are not allowed:

$$\frac{x}{x-1},\frac{x^3+2}{x^2+1},\frac{1}{x^2-1},\frac{1}{x(x-1)}$$

These are allowed, showing irreducible quadratic and higher powers:

$$\frac{x}{x^2+1},\frac{x^3+2x+1}{{(x^2+2)}^2}$$

In this example the numerator is linear and the denominator is quadratic and irreducible.

To create a partial fraction decomposition, follow these steps:

1. Check denominator degree is higher
   • Else do long division
2. Factor denominator completely (even using irrational roots)
3. Write the generic sum of partial fraction terms with their constants

Repeated factors – special treatment – incrementing powers

4. Solve for constants

Theory 2

Partial fractions can be integrated using just a few techniques. Consider these terms:

$$\begin{array}{c}\frac{A}{x-a},\frac{A}{{(x-a)}^2},\frac{A}{{(x-a)}^3},\dots ,\\ \\ \text{and}\frac{A}{x^2+h^2},\text{and}\frac{Ax+B}{x^2+h^2}\end{array}$$

Linear power bottom

In order to integrate terms like this:

$$\frac{A}{{(x-a)}^n}$$

If $n=1$ then use log:

$$\int \frac{A}{x-a}dx=A\ln |x-a|+C$$

If $n>1$ then use power rule:

$$\int \frac{A}{{(x-a)}^n}dx\gg \gg \int A{(x-a)}^{-n}dx\gg \gg A\frac{{(x-a)}^{-n+1}}{-n+1}+C$$

Quadratic bottom, constant top

Formula for simple irreducible quadratics:

$$\int \frac{dx}{x^2+h^2}=\frac{1}{h}{\tan }^{-1}\left(\frac{x}{h}\right)+C$$

Memorize this formula!

Quadratic bottom, linear top

In order to integrate terms like this:

$$\frac{Ax+B}{x^2+h^2}$$

Break into separate terms:

$$\frac{Ax+B}{x^2+h^2}\gg \gg \frac{Ax}{x^2+h^2}+\frac{B}{x^2+h^2}$$

Then:

• First term with $x$ in top:

$$\int \frac{Ax}{x^2+h^2}dx\gg \gg \frac{A}{2}\ln |x^2+h^2|+C$$

• Second term lacking $x$ in top:

$$\int \frac{B}{x^2+h^2}dx\gg \gg \frac{B}{h}{\tan }^{-1}\left(\frac{x}{h}\right)+C$$

Extra - Completing the square when “no real roots”

To integrate terms with more general quadratics, like this:

$$\frac{A}{x^2+bx+c}$$

we need $b^2-4c<0$, i.e. “no real roots” of the quadratic. If that holds, then we can complete the square and substitute $u=\tan \theta$ as follows.

Look what happens when completing the square:

$$x^2+bx+c\gg \gg {(x+\frac{b}{2})}^2-\frac{b^2}{4}+c$$

Notice that $b^2-4c<0$ is equivalent to the condition $-\frac{b^2}{4}+c>0$. Create a new label $Z=-\frac{b^2}{4}+c$. So this condition means $Z>0$ and we can safely define $\sqrt{Z}$.

Then a $u$-substitution $u=x+\frac{b}{2}$ simplifies the equation like this:

$$x^2+bx+c\gg \gg u^2+{\sqrt{Z}}^2$$

The quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ shows that the condition $b^2-4c<0$ is equivalent to the condition “no real roots.” (In our case $a=1$. If we had $a\ne 1$, we could divide out this $a$ and change the others.)

So we see that “no real roots” is equivalent to the condition that the denominator can be converted to the format $x^2+h^2$ with some constant $h$.

At this point, to compute the integral, do trig sub with $u=\sqrt{Z}\tan \theta$ and $du=\sqrt{Z}{\sec }^2\theta d\theta$:

$$\begin{array}{c}\int \frac{Adx}{x^2+bx+c}\gg \gg \int \frac{A\sqrt{Z}{\sec }^2\theta d\theta }{Z{\sec }^2\theta }\\ \\ \gg \gg \frac{A}{\sqrt{Z}}\int d\theta \gg \gg \frac{A}{\sqrt{Z}}\theta +C\\ \\ \gg \gg \frac{A}{\sqrt{Z}}{\tan }^{-1}\left(\frac{x+b/2}{\sqrt{Z}}\right)+C\end{array}$$

Extra - “Rationalize a quotient” - convert into PFD

Sometimes an integrand may be converted to a rational function using a substitution.

Consider this integral:

$$\int \frac{\sqrt{x+4}}{x}dx$$

Set $u=\sqrt{x+4}$, so $x=u^2-4$ and $dx=2udu$:

$$\gg \gg \int \frac{2udu}{u^2-4}$$

Now this rational function has a partial fraction decomposition:

$$\frac{2u}{u^2-4}\gg \gg \frac{2u}{(u-2)(u+2)}\gg \gg \frac{1}{u-2}+\frac{1}{u+2}$$

It is easy to integrate from there!

Practice exercises:

• To compute $\int \frac{\sqrt{x}}{x-1}dx$, try the substitution $u=\sqrt{x}$.
• To compute $\int \frac{dx}{\sqrt[2]{x}-\sqrt[3]{x}}$, try the substitution $u=\sqrt[6]{x}$.
• To compute $\int \frac{1}{x-\sqrt{x+2}}dx$, try the substitution $u=\sqrt{x+2}$.