Integrating by Long Division and… — AP® Calculus AB Notes

Substitution and the basic rules cover most integrals, but a few fractions need a little algebra first. This page covers the two rewrites on the AB exam: polynomial long division, and completing the square to reach an arctangent.

Long division: when the top is "too big"

The graph of y = (x squared + 3x + 5)/(x + 1) in two branches on either side of the vertical asymptote x = -1, with the dashed line y = x + 2. The curve hugs the line far from x = -1; at x = 3 a vertical segment marks the gap 3/(x + 1). Long division gives x + 2 + 3/(x + 1). — **Figure 1.** The fraction is the line y = x + 2 plus a small piece, 3/(x + 1). LevlPrep original diagram.

In $\frac{x^2 + 3x + 5}{x + 1}$ the top has degree 2 and the bottom degree 1. No basic rule fits, and substitution with $u = x + 1$ is clumsy. Divide, exactly as with numbers: $\frac{17}{5} = 3 + \frac{2}{5}$ .

Worked example 1. Find $\int \frac{x^2 + 3x + 5}{x + 1}\,dx$ .

Divide $x^2 + 3x + 5$ by $x + 1$ . First term: $x$ , since $x(x + 1) = x^2 + x$ . Subtract: $2x + 5$ remains.

Next term: $2$ , since $2(x + 1) = 2x + 2$ . Subtract: $3$ remains, the remainder.

So $\frac{x^2 + 3x + 5}{x + 1}$ $= x$ $+ 2$ $+ \frac{3}{x + 1}$ . (Check: $(x + 1)(x + 2) + 3$ $= x^2 + 3x + 5$ .)

Integrate each piece: $\frac{x^2}{2} + 2x + 3\ln|x + 1| + C$ .

The figure shows what the division means: far from $x = -1$ , the curve is nearly the line $y = x + 2$ , and the remainder $\frac{3}{x + 1}$ is the shrinking gap.

A quick version works when the bottom divides neatly into the top: $\frac{2x + 3}{x + 1}$ $= \frac{2(x + 1) + 1}{x + 1}$ $= 2$ $+ \frac{1}{x + 1}$ , and $\frac{x^2}{x^2 + 1}$ $= \frac{(x^2 + 1) - 1}{x^2 + 1}$ $= 1 - \frac{1}{x^2 + 1}$ .

Completing the square: reaching arctan

The rule $\int \frac{1}{1 + u^2}\,du$ $= \arctan u + C$ handles a sum of squares on the bottom. A quadratic with no real zeros can always be written as one. To complete the square in $x^2 + bx + c$ , take half of $b$ , square it, and balance:

$x^2 - 2x + 10 = (x^2 - 2x + 1) + 9 = (x - 1)^2 + 9.$

You also need the scaled version. With $u = x - h$ and $a > 0$ :

$\int \frac{1}{u^2 + a^2}\,du = \frac{1}{a}\arctan\frac{u}{a} + C.$

(Check by differentiating: $\frac{1}{a}\cdot\frac{1/a}{1 + u^2/a^2}$ $= \frac{1}{a^2 + u^2}$ .)

Worked example 2. Evaluate $\int_1^4 \frac{1}{x^2 - 2x + 10}\,dx$ .

Complete the square: $(x - 1)^2 + 9$ , so $u = x - 1$ and $a = 3$ .

Antiderivative: $\frac{1}{3}\arctan\frac{x - 1}{3}$ .

Evaluate: $\frac{1}{3}\left(\arctan 1 - \arctan 0\right)$ $= \frac{1}{3}\cdot\frac{\pi}{4}$ $= \frac{\pi}{12}$ .

Which rewrite, if any?

Integrand looks like	Do this	Example
top degree at least bottom degree	long division	$\frac{x^2 + 3x + 5}{x + 1}$ $= x$ $+ 2$ $+ \frac{3}{x + 1}$
top is the derivative of the bottom	substitute, get $\ln$	$\int \frac{2x - 2}{x^2 - 2x + 10}\,dx$ $= \ln(x^2 - 2x + 10)$ $+ C$
constant over a quadratic with no real zeros	complete the square, get arctan	$\int \frac{dx}{x^2 + 4x + 5}$ $= \arctan(x + 2)$ $+ C$

Two habits prevent most errors: never integrate the top and bottom of a fraction separately, and differentiate your answer to check it. The factor $\frac{1}{a}$ in the arctan form is the piece most often lost.

When the bottom factors into different linear pieces, such as $\frac{1}{(x - 1)(x + 2)}$ , the method of partial fractions splits it apart. That method is part of the BC course only and does not appear on the AB exam.