Integration Using Long Division (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

Integration using long division

How do I divide polynomials?

It is possible to use polynomial long division to simplify a rational function like $\frac{x^{3} + 6 x^{2} - 9 x - 11}{x - 2}$
- This can make the function much easier to integrate
Polynomial division works just like the long division method used for regular numbers:

2.5.2 Bus Stop Div, Edexcel A Level Maths: Pure revision notes

The answer to a polynomial division is built up term by term
- working downwards in powers of the variable (usually x)

E.g. dividing $x^{3} + 6 x^{2} - 9 x - 11$ by $x - 2$

Beginning of a polynomial long division problem, dividing x^3+6x^2-9x-11 by x-2

$x^{3} + 6 x^{2} - 9 x - 11$ (the thing being divided) is known as the dividend
- and $x - 2$ (the thing we're dividing by) is known as the divisor
Start by dealing with the highest power term in the dividend ( $x^{3}$ )
- Compare the highest power term in the divisor ( $x$ )
- $x^{3} \div x = x^{2}$ so
  - put $x^{2}$ on top of the division line
  - and subtract $x^{2} \cdot (x - 2) = x^{3} - 2 x^{2}$ from the dividend

First step of a polynomial long division problem, dividing x^3+6x^2-9x-11 by x-2

Now deal with the highest power remaining in the expression on the bottom line ( $8 x^{2}$ )
- Compare the highest power term in the divisor ( $x$ )
- $8 x^{2} \div x = 8 x$ so
  - add $8 x$ on top of the division line
  - and subtract $8 x \cdot (x - 2) = 8 x^{2} - 16 x$ from the bottom line

Second step of a polynomial long division problem, dividing x^3+6x^2-9x-11 by x-2

Now deal with the highest power remaining in the expression on the bottom line ( $7 x$ )
- Compare the highest power term in the divisor ( $x$ )
- $7 x \div x = 7$ so
  - add $7$ on top of the division line
  - and subtract $7 \cdot (x - 2) = 7 x - 14$ from the bottom line

Third step of a polynomial long division problem, dividing x^3+6x^2-9x-11 by x-2

The 3 'left over' at the bottom is the remainder
- Therefore $\frac{x^{3} + 6 x^{2} - 9 x - 11}{x - 2} = x^{2} + 8 x + 7 + \frac{3}{x - 2}$
- In that new form, the function would be very easy to integrate

Exam Tip

Be extra careful when subtracting expressions with negative coefficients

Using brackets can help you keep track of things

Worked Example

Find the indefinite integral $\int \frac{x^{3} + 2 x + 13}{x + 3} d x$ .

Answer:

Start by using polynomial long division to rewrite the function being integrated

Add in $+ 0 x^{2}$ to the dividend as a placeholder for the 'missing' $x^{2}$ term

The workings out for a polynomial long division problem, dividing x^3++2x+13 by x+3

That means $\frac{x^{3} + 2 x + 13}{x + 3} = x^{2} - 3 x + 11 - \frac{20}{x + 3}$ , which may be integrated easily

$\begin{array}{rcl} \int \frac{x^{3} + 2 x + 13}{x + 3} d x & = & \int (x^{2} - 3 x + 11 - \frac{20}{x + 3}) d x \\ = & \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 11 x - 20 \ln |x + 3| + C \end{array}$

$\begin{array}{rcl} \int \frac{x^{3} + 2 x + 13}{x + 3} d x & = & \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 11 x - 20 \ln |x + 3| + C \end{array}$

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