Divide or cancel algebraic factors

A fraction that splits

The chalk curve is \(\dfrac{x^2+3x+4}{x+2}\). The degree of a polynomial is its highest power of \(x\), so the top has degree 2 and the bottom has degree 1. No basic rule gives its antiderivative.

Divide the top by the bottom, as in long division of numbers. The quotient is \(x + 1\) and the remainder is 2, because \((x+2)(x+1) + 2 = x^2+3x+4\).

Divide both sides by \(x + 2\). The curve equals \(x + 1\) plus \(\dfrac{2}{x+2}\). Press the button, and the orange dashed curve draws this sum over the chalk curve.

Drag the orange line to \(x = 2\). The curve has height 3.5 there, the line \(y = x + 1\) has height 3, and the gap between them is \(\tfrac24 = 0.5\).

The region from \(x = 0\) to \(x = 2\) splits in two. The teal region under the line has area 4. The green region between the line and the curve has height \(\dfrac{2}{x+2}\), so its area is \(2\ln 2\), about 1.39.

Any fraction of polynomials

The curve is a fraction of two polynomials. If the degree of the top is at least the degree of the bottom, divide the top by the bottom. The division gives a quotient and a remainder of lower degree than the bottom.

Move the Remainder slider. The teal region under the straight line, the quotient, stays in place. The green region between the line and the curve grows or shrinks, because its height is the remainder over the bottom.

The dashed vertical line stands where the bottom is 0, so the fraction has no value there. With a remainder, the curve rises or falls without end beside this line, which is an asymptote.

Dividing changes the formula, not the curve. The sum has the same height as the fraction at every \(x\), so you can integrate each term.

Notation

1. Write \(P\) for the top, \(Q\) for the bottom, \(S\) for the quotient and \(R\) for the remainder.

   \[P(x) = Q(x)\,\cb{S(x)} + R(x)\]

2. Divide both sides by \(Q\). The fraction equals the quotient plus the remainder over the bottom.

   \[\dfrac{P(x)}{Q(x)} = \cb{S(x)} + \cg{\dfrac{R(x)}{Q(x)}}\]

3. Integrate each term. A constant \(c\) over a linear bottom \(ax + b\) gives a logarithm, because the substitution \(u = ax + b\) gives \(du = a\,dx\).

   \[\int \dfrac{c}{ax+b}\,dx = \dfrac{c}{a}\ln|ax+b| + C\]

4. If the top and the bottom share a factor, cancel it. The fraction has no value where that factor is 0, so write the excluded value with the result.

   \[\dfrac{(x-2)(x+2)}{x-2} = x + 2, \quad x \neq 2\]

When it fails

In this figure the bottom \(Q\) is \(x + 2\), which is 0 at \(x = -2\), so the curve has no value there.

A region that crosses \(x = -2\) has no area that these rules can find. Keep each integral on one side.

Example 1: cancel a factor

Evaluate \(\displaystyle\int_0^3 \frac{x^2-25}{x+5}\,dx\).

1. Factor the top as a difference of squares. The factor \(x + 5\) is also the bottom.

   \[x^2 - 25 = (x-5)(x+5)\]

2. Cancel the common factor. The bottom is 0 at \(x = -5\), so the original fraction has no value there and the result holds only for \(x \neq -5\). The open circle marks the missing point.

   \[\dfrac{x^2-25}{x+5} = x - 5, \quad x \neq -5\]

3. The interval from 0 to 3 does not contain \(-5\), so the open circle does not affect the integral. The line lies below the x-axis there, so the red region counts as negative.

   \[\begin{aligned} \int_0^3 (x-5)\,dx &= \left[\dfrac{x^2}{2} - 5x\right]_0^3 \\ &= \dfrac92 - 15 = \cr{-\dfrac{21}{2}} \end{aligned}\]

Example 2: divide

Evaluate \(\displaystyle\int_1^2 \frac{4x^2+4x+5}{2x-1}\,dx\).

1. The top has degree 2 and the bottom has degree 1, so divide. The quotient is \(2x + 3\) and the remainder is 8.

   \[\begin{aligned} 4x^2+4x+5 &= (2x-1)(2x+3) \\ &\quad + 8 \end{aligned}\]

2. Divide both sides by \(2x - 1\).

   \[\begin{aligned} \dfrac{4x^2+4x+5}{2x-1} &= \cb{2x+3} \\ &\quad + \cg{\dfrac{8}{2x-1}} \end{aligned}\]

3. Integrate the quotient with the power rule. The teal region under the line has area 6.

   \[\int_1^2 (2x+3)\,dx = \left[x^2 + 3x\right]_1^2 = \cb{6}\]

4. A tempting wrong move: Integrate the remainder term as 8 times the logarithm of the bottom.

   \[\int \dfrac{8}{2x-1}\,dx = \cr{8\ln|2x-1|}\]

   From \(x = 1\) to \(x = 2\) this gives \(8\ln 3 - 8\ln 1 = 8\ln 3\), about 8.79. But the term \(\dfrac{8}{2x-1}\) is at most 8 tall and 1 wide, so the green region has area at most 8. The slope of \(8\ln|2x-1|\) is \(\dfrac{16}{2x-1}\), twice the term, so the red curve, the line plus this slope, lies far above the chalk curve.

5. Let \(u = 2x - 1\). Then \(dx = \tfrac12\,du\), which halves the coefficient, so the green region has area \(4\ln 3\).

   \[\int \dfrac{8}{2x-1}\,dx = \cg{4\ln|2x-1|}\]

6. Add the two areas. As a check, the value is about 10.39.

   \[\begin{aligned} &\left[x^2 + 3x + 4\ln|2x-1|\right]_1^2 \\ &\quad = 6 + 4\ln 3 \end{aligned}\]

Practice

Evaluate \(\displaystyle\int_0^2 \frac{x^2+5x+1}{x+3}\,dx\).

Divide, because the top has the higher degree. The quotient is a straight line, and the remainder term is the gap from that line to the curve.

Questions

The chalk curve is \(y = \dfrac{x^2+2}{x-1}\). The faint dashed line is \(y = x + 1\), the quotient when \(x^2 + 2\) is divided by \(x - 1\). The remainder is 3.

1. At \(x = 4\) the curve is 1 above the faint line. Which expression equals this gap?

   • \(\dfrac{3}{x-1}\)
   • 3
   • \(x + 1\)
   Answer

   \(\dfrac{3}{x-1}\)

2. Which interval can you integrate over with these rules?

   • From 2 to 5
   • From 0 to 3
   • From \(-2\) to 4
   Answer

   From 2 to 5

3. Evaluate \(\displaystyle\int_3^5 \frac{x^2-4}{x-2}\,dx\).

   • 12
   • 8
   • 20
   Answer

   12

Summary

You divide when the degree of the top is at least the degree of the bottom, and you cancel a shared factor.

This works because the new form equals the fraction wherever the fraction has a value, so you can integrate each term.

You keep each integral on one side of any x where the bottom is 0.

This is one of four skills in topic 6.10 of Which & Why: AP Calculus.