Choose an algebraic rewrite
One curve, two formulas
The chalk curve is \(f(x) = \dfrac{(2x+1)^2}{x}\), for \(x > 0\). No row of the table of antiderivatives matches this formula.
Move the Trace slider. The orange dashed curve is drawn from the formula \(4x + 4 + \tfrac1x\). It lies on the chalk curve everywhere, so the two formulas describe the same function.
You find the new formula in two steps. Squaring the top gives \(4x^2 + 4x + 1\), and dividing each term by \(x\) gives \(4x + 4 + \tfrac1x\).
Move the Split slider. The region from \(x = 1\) to \(x = 2\) splits into three bands, one for each term. The teal band is the area under \(4x\), and each green band adds the area of the next term, \(4\) and then \(\tfrac1x\).
The rewrite did not change the area, but it changed the integrand into terms that you can integrate.
Same function, easier terms
Press each button, or move the Trace slider. The orange dashed curve lies on the chalk curve every time, so a rewrite does not change the function.
The goal of a rewrite is to give terms that each match a row of the table, or each fit a substitution.
The choice of the rewrite depends on the shape of the integrand.
Notation
A rewrite replaces the integrand by a formula with the same value at every \(x\). Move the \(x\) slider. The orange line reaches both curves at the same height, so the integral does not change.
If the integrand is a product or a power of a sum, expand it.
\[(x+1)^2 = x^2 + 2x + 1\]If the numerator of a fraction has at least the degree of the denominator, divide. The quotient \(Q\) is a polynomial, and the remainder \(R\) has a lower degree than \(D\).
\[\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}\]If the numerator is a sum and the denominator has one term, split the fraction.
\[\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}\]If the integrand contains a trigonometric sum, an identity can turn it into one term.
\[1 + \cos x = 2\cos^2\frac{x}{2}\]If a quadratic does not fit a row, complete the square.
\[\begin{aligned} &x^2 + bx + c \\ &\quad = \left(x + \frac{b}{2}\right)^2 + c - \frac{b^2}{4} \end{aligned}\]A rewrite is good when every term matches a row or a substitution, so each band of the region is an area you can find. Otherwise, choose another rewrite.
When it fails
This rewrite cancels the factor \(x - 1\), which is 0 at \(x = 1\). There the original has no value, because its denominator is 0, but the new formula has the value 2.
A single point has no area, so both formulas give the same integral over an interval. The rewrite is valid for \(x \ne 1\).
Example 1: divide first
Find \(\displaystyle\int \frac{x^2 + 5}{x - 1}\,dx\), for \(x > 1\).
No row matches, and the numerator has degree 2 while the denominator has degree 1, so divide. The quotient is \(x + 1\) and the remainder is 6.
\[x^2 + 5 = (x - 1)(x + 1) + 6\]Divide both sides by \(x - 1\). The orange dashed curve lies on the chalk curve, so the two formulas are the same.
\[\frac{x^2 + 5}{x - 1} = x + 1 + \frac{6}{x - 1}\]Integrate each term. The first two terms are powers. The third term is the row for \(\tfrac1x\) with \(x - 1\) in place of \(x\), because the derivative of \(x - 1\) is 1.
\[\frac{x^2}{2} + x + 6\ln|x - 1| + C\]
Example 2: a rewrite that fits a row
Find \(\displaystyle\int \frac{dx}{1 + \cos x}\), for \(-\pi < x < \pi\).
No row matches \(\tfrac{1}{1 + \cos x}\), and the denominator is a sum of two terms, so rewrite it. At \(x = 0\) the value is \(\tfrac{1}{1+1} = \tfrac12\), which the orange line marks.
A tempting wrong move: Split the fraction into one fraction for each term of the denominator.
\[\frac{1}{1 + \cos x} = 1 + \sec x\]Test the new formula at \(x = 0\). It gives \(1 + 1 = 2\), but the original gives \(\tfrac12\). In the picture, the red curve \(1 + \sec x\) lies above the chalk curve.
Use an identity instead. Because \(1 + \cos x = 2\cos^2\tfrac{x}{2}\), the integrand is a multiple of \(\sec^2\), and the orange dashed curve lies on the chalk curve.
\[\begin{aligned} &\frac{1}{1 + \cos x} \\ &\quad = \frac{1}{2\cos^2(x/2)} \\ &\quad = \tfrac12\sec^2\frac{x}{2} \end{aligned}\]The row for \(\sec^2\) gives \(\tan\). The inner function \(\tfrac{x}{2}\) has the constant derivative \(\tfrac12\), which is already a factor of the integrand.
\[\begin{aligned} &\int \tfrac12\sec^2\frac{x}{2}\,dx \\ &\quad = \tan\frac{x}{2} + C \end{aligned}\]
Practice
Find \(\displaystyle\int \frac{dx}{x^2 + 6x + 13}\).
The peak of the bell is at \(x = -3\), so the rewritten denominator should contain \(x + 3\). Compare it with the row for \(\tfrac{1}{u^2 + a^2}\).
Solution
The denominator does not factor, and its derivative \(2x + 6\) is not in the numerator, so no row and no substitution fits. Complete the square. The orange dashed curve lies on the chalk curve.
\[x^2 + 6x + 13 = (x + 3)^2 + 4\]The integrand now matches the row for \(\tfrac{1}{u^2 + a^2}\), with \(u = x + 3\) and \(a = 2\).
\[\begin{aligned} &\int \frac{dx}{(x+3)^2 + 2^2} \\ &\quad = \frac12\arctan\frac{x+3}{2} + C \end{aligned}\]
Questions
The chalk curve is \(f(x) = \dfrac{x^2 + 2x - 3}{x - 1}\), and the open circle shows where \(f\) has no value.
Which formula is equal to \(f(x)\) for every \(x \ne 1\)?
- \(x + 3\)
- \(x^2 + 2x - 3\)
- \(x - 3\)
Answer
\(x + 3\)
What is the area of the region under the curve from \(x = 0\) to \(x = 2\)?
- 8
- 6
- The area does not exist, because \(f(1)\) has no value
Answer
8
Which function is \(\displaystyle\int \frac{x^3 + 2x}{x^2}\,dx\)?
- \(\dfrac{x^2}{2} + 2\ln|x| + C\)
- \(\dfrac{x^4/4 + x^2}{x^3/3} + C\)
- \(\dfrac{x^2}{2} + 2x + C\)
Answer
\(\dfrac{x^2}{2} + 2\ln|x| + C\)
Summary
You change the integrand by expanding, dividing, splitting, using an identity or completing the square, until each term matches a row or a substitution.
This works because a rewrite gives the same function, so the integral does not change, and the new terms are easier to integrate.
You use a rewrite when no row and no inner function fit the integrand as it is.
This is one of five skills in topic 6.14 of Which & Why: AP Calculus.