Calculus 1

Dividing before integrating

The table of antiderivatives has a line for one over x and a line for one over one plus x squared, but nothing for a fraction whose top is a bigger polynomial than its bottom. How can such a fraction be turned into pieces the table already knows?

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Antiderivatives of trig, exponential and log functionsCalculus 1 Polynomial long divisionAlgebra 2

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Integration by partial fractionsCalculus 2

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Dividing before integrating

Improper integrals with an infinite integrand Integration by parts ▾ Show all 16 (4 more)

When you reach for it

A rational function is a quotient \dfrac{P(x)}{Q(x)} of two polynomials. Call it top-heavy when the degree of P is at least the degree of Q, and proper when the degree of P is less. An integral such as

\int \frac{x^3 + x^2 + 4x + 9}{x^2 + 4}\,dx

matches no line of the table in Antiderivatives of trig, exponential and log functions, and splitting it term by term over the denominator, which worked for a one-term denominator such as x^2, leaves fractions like \dfrac{x^3}{x^2 + 4} that are no better. What does work is the move from arithmetic: an improper fraction such as \dfrac{17}{5} is 3 + \dfrac{2}{5}, a whole number plus a proper fraction. Dividing the polynomials does the same thing, and leaves a polynomial, which the power rule integrates, plus a proper remainder fraction, which for the divisors in this lesson is a logarithm or an arctangent.

Reach for division whenever the integrand is a top-heavy rational function. A proper integrand needs no division; it is already its own remainder.

The procedure

Dividing first

Polynomial long division gives, for polynomials P and Q with Q not the zero polynomial, a quotient S and a remainder R with

\begin{aligned} &P(x) = Q(x)\,S(x) + R(x), \\ &R = 0 \text{ or } \deg R < \deg Q. \end{aligned}

Dividing both sides by Q(x), wherever Q(x) \ne 0,

\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}.

This is an identity: the two sides are the same function on every interval where Q has no zero, so they have the same antiderivatives there. The sum rule then lets us integrate S and \dfrac{R}{Q} separately. When the divisor is x - r, Synthetic division finds S and R with less writing, and the remainder is the number P(r).

A remainder over a linear divisor

Dividing by ax + b with a \ne 0 leaves a constant remainder k, so the leftover piece is \dfrac{k}{ax + b}. On each interval where ax + b \ne 0,

\int \frac{1}{ax + b}\,dx = \frac{1}{a}\ln\lvert ax + b \rvert + C.

Why. The derivative of \ln\lvert u \rvert is \dfrac{1}{u} for u \ne 0, and the chain rule with u = ax + b gives

\frac{d}{dx}\Big[\ln\lvert ax + b \rvert\Big] = \frac{1}{ax + b} \cdot a = \frac{a}{ax + b}.

That is a times too big, so dividing by a gives an antiderivative. When a = 1 the factor disappears: \displaystyle \displaystyle\int \frac{1}{x + c}\,dx = \ln\lvert x + c \rvert + C. This is the first case of u-substitution, which handles any inside function; here the inside is linear and differentiating the answer is all the check we need.

A remainder over x^2 + c^2

The divisor x^2 + c^2 with c > 0 is never zero. Dividing by it leaves a remainder of degree at most 1, say px + q, and the remainder fraction splits over its single denominator:

\frac{px + q}{x^2 + c^2} = p \cdot \frac{x}{x^2 + c^2} + q \cdot \frac{1}{x^2 + c^2}.

Each piece has an antiderivative on all of \mathbb{R}:

\int \frac{x}{x^2 + c^2}\,dx = \frac{1}{2}\ln\left(x^2 + c^2\right) + C,

\int \frac{1}{x^2 + c^2}\,dx = \frac{1}{c}\arctan\frac{x}{c} + C.

Why. By the chain rule, \dfrac{d}{dx}\Big[\ln\left(x^2 + c^2\right)\Big] = \dfrac{2x}{x^2 + c^2}, so half of it is an antiderivative of \dfrac{x}{x^2 + c^2}. No absolute value is needed, because x^2 + c^2 > 0. For the second, since \dfrac{d}{dx}\big[\arctan x\big] = \dfrac{1}{1 + x^2},

\frac{d}{dx}\left[\frac{1}{c}\arctan\frac{x}{c}\right] = \frac{1}{c} \cdot \frac{1}{1 + \frac{x^2}{c^2}} \cdot \frac{1}{c} = \frac{1}{c^2 + x^2}.

With c = 1 this is the table’s \int \dfrac{1}{1 + x^2}\,dx = \arctan x + C.

Definite integrals

To find \int_a^b \dfrac{P(x)}{Q(x)}\,dx, divide, find an antiderivative F, and use F(b) - F(a) from The second fundamental theorem of calculus. That theorem needs the integrand continuous on [a, b], so Q must have no zero in [a, b]. A zero of Q that is not also a zero of P is a vertical asymptote, where the integrand is unbounded and the integral does not exist as an ordinary integral at all. When the interval lies to the left of a zero r of x - r, the absolute value matters: there \lvert x - r \rvert = r - x.

Worked examples

Example 1

Synthetic division by x + 1

Find \int \dfrac{2x^3 + x^2 - 4x + 1}{x + 1}\,dx.

Solution

The top has degree 3 and the bottom degree 1, so divide. The divisor is x - (-1), so synthetic division uses r = -1 on the coefficients 2, 1, -4, 1. Bring down 2. Then 2 \cdot (-1) = -2 and 1 + (-2) = -1; next (-1)(-1) = 1 and -4 + 1 = -3; last (-3)(-1) = 3 and 1 + 3 = 4. The quotient is 2x^2 - x - 3 and the remainder is 4:

\frac{2x^3 + x^2 - 4x + 1}{x + 1} = 2x^2 - x - 3 + \frac{4}{x + 1}.

Check the division: (x + 1)(2x^2 - x - 3) + 4 = 2x^3 + x^2 - 4x - 3 + 4, the original top. The polynomial integrates by the power rule and the last term by the linear formula with a = 1:

\begin{aligned} &\int \frac{2x^3 + x^2 - 4x + 1}{x + 1}\,dx \\ &\quad = \frac{2x^3}{3} - \frac{x^2}{2} - 3x + 4\ln\lvert x + 1 \rvert + C. \end{aligned}

The answer holds on (-\infty, -1) and on (-1, \infty) separately, the two intervals where the integrand is defined.

Example 2

Malu checks the factor of one half

Find \int \dfrac{6x + 1}{2x - 1}\,dx.

Solution

Top and bottom both have degree 1, so the integrand is top-heavy. Since 3(2x - 1) = 6x - 3, the top is 6x + 1 = 3(2x - 1) + 4, and

\frac{6x + 1}{2x - 1} = 3 + \frac{4}{2x - 1}.

Malu’s first guess for the second piece is 4\ln\lvert 2x - 1 \rvert. She differentiates it: the chain rule gives 4 \cdot \dfrac{2}{2x - 1} = \dfrac{8}{2x - 1}, twice the integrand. Halving the guess fixes it, which is the factor \dfrac{1}{a} = \dfrac{1}{2} in the formula. So

\int \frac{6x + 1}{2x - 1}\,dx = 3x + 2\ln\lvert 2x - 1 \rvert + C,

on \displaystyle x < \frac{1}{2} and on \displaystyle x > \frac{1}{2}. Check: the derivative is 3 + \dfrac{4}{2x - 1} = \dfrac{6x - 3 + 4}{2x - 1}.

Example 3

A remainder that becomes an arctangent

Find \int \dfrac{x^3 + x^2 + 4x + 9}{x^2 + 4}\,dx.

Solution

Long division by x^2 + 4: the first term of the quotient is \dfrac{x^3}{x^2} = x, and subtracting x(x^2 + 4) = x^3 + 4x leaves x^2 + 9. The next term is 1, and subtracting x^2 + 4 leaves 5, which has degree 0 < 2, so the division stops:

\frac{x^3 + x^2 + 4x + 9}{x^2 + 4} = x + 1 + \frac{5}{x^2 + 4}.

The remainder fraction is 5 times the form \dfrac{1}{x^2 + c^2} with c = 2, whose antiderivative is \dfrac{1}{2}\arctan\dfrac{x}{2}. So

\begin{aligned} &\int \frac{x^3 + x^2 + 4x + 9}{x^2 + 4}\,dx \\ &\quad = \frac{x^2}{2} + x + \frac{5}{2}\arctan\frac{x}{2} + C, \end{aligned}

valid for every real x, since x^2 + 4 is never zero.

Example 4

Nukappi’s remainder needs both forms

Find \int \dfrac{x^3 + 2x^2 + 3}{x^2 + 1}\,dx.

Solution

Nukappi writes the top as x^3 + 2x^2 + 0x + 3, with a placeholder for the missing power. Subtracting x(x^2 + 1) = x^3 + x leaves 2x^2 - x + 3; subtracting 2(x^2 + 1) leaves -x + 1, of degree 1 < 2. So

\begin{aligned} \frac{x^3 + 2x^2 + 3}{x^2 + 1} &= x + 2 + \frac{1 - x}{x^2 + 1} \\ &= x + 2 + \frac{1}{x^2 + 1} - \frac{x}{x^2 + 1}. \end{aligned}

The remainder has both a constant and an x on top, so it splits into an arctangent piece and a logarithm piece, each with c = 1:

\begin{aligned} &\int \frac{x^3 + 2x^2 + 3}{x^2 + 1}\,dx \\ &\quad = \frac{x^2}{2} + 2x + \arctan x - \frac{1}{2}\ln\left(x^2 + 1\right) + C. \end{aligned}

Nukappi checks by differentiating: x + 2 + \dfrac{1}{1 + x^2} - \dfrac{1}{2} \cdot \dfrac{2x}{x^2 + 1}, which is the divided form of the integrand.

Example 5

A definite integral, and the slant asymptote

Evaluate \int_2^4 \dfrac{x^2 + x - 1}{x - 1}\,dx.

Solution

Synthetic division with r = 1 on 1, 1, -1: bring down 1; 1 + 1 = 2; -1 + 2 = 1. So

\frac{x^2 + x - 1}{x - 1} = x + 2 + \frac{1}{x - 1}.

The only zero of the bottom is 1, outside [2, 4], so the integrand is continuous there and the theorem applies. On [2, 4], x - 1 > 0, so F(x) = \dfrac{x^2}{2} + 2x + \ln(x - 1) is an antiderivative, and

\begin{aligned} \int_2^4 \frac{x^2 + x - 1}{x - 1}\,dx &= F(4) - F(2) \\ &= (8 + 8 + \ln 3) - (2 + 4 + \ln 1) \\ &= 10 + \ln 3 \approx 11.099. \end{aligned}

Figure 1 reads the answer. The quotient x + 2 is the slant asymptote of the graph, and \int_2^4 (x + 2)\,dx = 10 is the trapezoid under it. The remainder \dfrac{1}{x - 1} is the gap between curve and line, and its integral \ln 3 is the area of that strip. As a rough check, the integrand runs from 5 to \displaystyle \frac{19}{3} on an interval of width 2, so the integral lies between 10 and about 12.7.

Figure 1Example 5. After division, \displaystyle \frac{x^2 + x - 1}{x - 1} = x + 2 + \frac{1}{x - 1}. The shaded area is the area 10 under the line y = x + 2 plus the area \ln 3 of the thin strip between the curve and the line.

Example 6

Sakari integrates across an asymptote

Sakari writes

\begin{aligned} \int_0^3 \frac{x^2}{x - 2}\,dx &= \left(\frac{x^2}{2} + 2x + 4\ln\lvert x - 2 \rvert\right)\Big|_0^3 \\ &= \frac{21}{2} - 4\ln 2. \end{aligned}

Her division \dfrac{x^2}{x - 2} = x + 2 + \dfrac{4}{x - 2} and her antiderivative are both right. Explain what is wrong, and evaluate \int_{-1}^{1} \dfrac{x^2}{x - 2}\,dx instead.

Solution

The point x = 2 lies inside [0, 3]. There the bottom is 0 and the top is 4, so the integrand is unbounded near 2: a vertical asymptote. The theorem F(b) - F(a) needs the integrand continuous on the whole interval, and F itself is not even defined at 2, so the computation proves nothing. In fact, as x \to 2 the term 4\ln\lvert x - 2 \rvert falls without bound, the Riemann sums have no limit, and \int_0^3 \dfrac{x^2}{x - 2}\,dx does not exist. Integrals of this kind are studied as improper integrals.

On [-1, 1] the bottom x - 2 is negative and never zero, so the theorem applies, with \lvert x - 2 \rvert = 2 - x:

\begin{aligned} F(1) &= \frac{1}{2} + 2 + 4\ln 1 = \frac{5}{2}, \\ F(-1) &= \frac{1}{2} - 2 + 4\ln 3 = -\frac{3}{2} + 4\ln 3, \\ F(1) - F(-1) &= 4 - 4\ln 3 \approx -0.394. \end{aligned}

The answer is negative, as it must be: on [-1, 1] the top x^2 is never negative and the bottom is always negative.

Boundaries

Division helps only a top-heavy integrand. A proper fraction divides to quotient 0 and leaves itself as the remainder.
Two remainder forms are covered here: a constant over ax + b, and px + q over x^2 + c^2. A remainder over a quadratic with real zeros, such as \dfrac{1}{x^2 - 1}, or over a product of factors, needs partial fractions. A quadratic with a middle term, such as x^2 + 2x + 5, is first reshaped by completing the square.
Each antiderivative holds on an interval where Q(x) \ne 0. On two separate intervals the constants C need not be equal.
A definite integral needs [a, b] free of zeros of Q. Across a vertical asymptote the integral does not exist, and F(b) - F(a) is a meaningless number.
The linear formula needs a \ne 0. With a = 0 the divisor is a constant and there is nothing to divide.

Where this goes next

u-substitution turns the check-by-differentiating used here for ax + b and x^2 + c^2 into a method for any inside function. Partial fractions picks up exactly where division stops: it splits a proper remainder over a factored denominator into pieces of the two forms above. Completing the square inside an integral widens the arctangent form, and Choosing an integration technique gathers these moves into one way of reading an integrand.

Exercises

Linear divisors

Find \int \dfrac{x^2 + 5x + 7}{x + 2}\,dx.
Find \int \dfrac{x^3 - 8}{x - 2}\,dx.
Find \int \dfrac{3x + 1}{x + 4}\,dx.
Find \int \dfrac{x^3 - 2x + 5}{x - 3}\,dx.
Find \int \dfrac{8x + 5}{4x + 1}\,dx.
Find \int \dfrac{6x^2 + 5x}{3x - 2}\,dx.
Find \int \dfrac{x^4}{x + 1}\,dx.

Quadratic divisors

Find \int \dfrac{2x^2 + 3}{x^2 + 9}\,dx.
Find \int \dfrac{x^3 + 5x}{x^2 + 4}\,dx.
Find \int \dfrac{x^3 - x^2 + 2}{x^2 + 1}\,dx.
Find \int \dfrac{x^4 + 5}{x^2 + 1}\,dx.
Find \int \dfrac{x^3 + 2x^2 + 7x + 10}{x^2 + 9}\,dx.

Definite integrals

Evaluate \int_0^2 \dfrac{x^2 + 2x + 3}{x + 1}\,dx.
Evaluate \int_3^5 \dfrac{2x - 1}{x - 2}\,dx.
Evaluate \int_0^2 \dfrac{x^2 + 8}{x^2 + 4}\,dx.
Evaluate \int_{-2}^{0} \dfrac{x^2 - 3}{x - 1}\,dx.
Evaluate \int_0^1 \dfrac{x^3 + x + 1}{x^2 + 1}\,dx.
Evaluate \int_0^2 \dfrac{x^3 + 3x + 2}{x^2 + 4}\,dx.

Reasoning

Explain why \int_0^4 \dfrac{x^2 + 4}{x - 3}\,dx cannot be found as F(4) - F(0) for an antiderivative F obtained by division.
Prove that \int \dfrac{1}{ax + b}\,dx = \dfrac{1}{a}\ln\lvert ax + b \rvert + C on each interval where ax + b \ne 0, for constants a \ne 0 and b.
Explain why the antiderivative of \dfrac{1}{x - c} needs an absolute value but the antiderivative \dfrac{1}{2}\ln\left(x^2 + c^2\right) of \dfrac{x}{x^2 + c^2}, with c > 0, does not.
Find the constant k for which \int \dfrac{x^2 + kx}{x - 2}\,dx has no logarithm in it, and give the antiderivative for that k.
Let g(x) = \dfrac{x^2 + 3x + 1}{x + 1}. Use division to show that for x \ge 0 the graph of g lies below the line y = x + 2 by a gap that shrinks toward 0, yet the area between them from 0 to b grows without bound as b \to \infty.

Answers

\dfrac{x^2}{2} + 3x + \ln\lvert x + 2 \rvert + C. Division: x^2 + 5x + 7 = (x + 2)(x + 3) + 1.
\dfrac{x^3}{3} + x^2 + 4x + C, for x \ne 2. The remainder is 0: x^3 - 8 = (x - 2)(x^2 + 2x + 4), so no logarithm appears.
3x - 11\ln\lvert x + 4 \rvert + C. Division: 3x + 1 = 3(x + 4) - 11.
\dfrac{x^3}{3} + \dfrac{3x^2}{2} + 7x + 26\ln\lvert x - 3 \rvert + C. Synthetic division with r = 3 on 1, 0, -2, 5 gives 1, 3, 7 and remainder 26.
2x + \dfrac{3}{4}\ln\lvert 4x + 1 \rvert + C. Division: 8x + 5 = 2(4x + 1) + 3, and a = 4 gives the factor \dfrac{1}{4}.
x^2 + 3x + 2\ln\lvert 3x - 2 \rvert + C. Division: 6x^2 + 5x = (3x - 2)(2x + 3) + 6, and \dfrac{6}{3} = 2.
\dfrac{x^4}{4} - \dfrac{x^3}{3} + \dfrac{x^2}{2} - x + \ln\lvert x + 1 \rvert + C. Synthetic division with r = -1 on 1, 0, 0, 0, 0 gives 1, -1, 1, -1 and remainder 1.
2x - 5\arctan\dfrac{x}{3} + C. Division: 2x^2 + 3 = 2(x^2 + 9) - 15, and \dfrac{15}{3} = 5.
\dfrac{x^2}{2} + \dfrac{1}{2}\ln\left(x^2 + 4\right) + C. Division: x^3 + 5x = x(x^2 + 4) + x.
\dfrac{x^2}{2} - x + 3\arctan x - \dfrac{1}{2}\ln\left(x^2 + 1\right) + C. Division: quotient x - 1, remainder 3 - x.
\dfrac{x^3}{3} - x + 6\arctan x + C. Division: x^4 + 5 = (x^2 + 1)(x^2 - 1) + 6.
\dfrac{x^2}{2} + 2x - \ln\left(x^2 + 9\right) - \dfrac{8}{3}\arctan\dfrac{x}{3} + C. Division: quotient x + 2, remainder -2x - 8.
4 + 2\ln 3. The integrand is x + 1 + \dfrac{2}{x + 1}, and \left(\dfrac{x^2}{2} + x + 2\ln(x + 1)\right)\Big|_0^2 = 4 + 2\ln 3.
4 + 3\ln 3. The integrand is 2 + \dfrac{3}{x - 2}, and \big(2x + 3\ln(x - 2)\big)\Big|_3^5 = (10 + 3\ln 3) - 6.
2 + \dfrac{\pi}{2}. The integrand is 1 + \dfrac{4}{x^2 + 4}, and \left(x + 2\arctan\dfrac{x}{2}\right)\Big|_0^2 = 2 + 2 \cdot \dfrac{\pi}{4}.
2\ln 3. The integrand is x + 1 - \dfrac{2}{x - 1}; on [-2, 0], \lvert x - 1 \rvert = 1 - x, and \left(\dfrac{x^2}{2} + x - 2\ln(1 - x)\right)\Big|_{-2}^{0} = 0 - (-2\ln 3).
\dfrac{1}{2} + \dfrac{\pi}{4}. The integrand is x + \dfrac{1}{x^2 + 1}, and \left(\dfrac{x^2}{2} + \arctan x\right)\Big|_0^1.
2 + \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2. The integrand is x + \dfrac{2}{x^2 + 4} - \dfrac{x}{x^2 + 4}, with antiderivative \dfrac{x^2}{2} + \arctan\dfrac{x}{2} - \dfrac{1}{2}\ln\left(x^2 + 4\right); its values are 2 + \dfrac{\pi}{4} - \dfrac{1}{2}\ln 8 at 2 and -\dfrac{1}{2}\ln 4 at 0.
Division gives x + 3 + \dfrac{13}{x - 3}, so F(x) = \dfrac{x^2}{2} + 3x + 13\ln\lvert x - 3 \rvert. The point 3 is inside [0, 4], where the bottom is 0 and the top is 13: the integrand is unbounded there, not continuous on [0, 4], and F is undefined at 3. The theorem does not apply, and the integral does not exist.
For ax + b > 0, \dfrac{d}{dx}\Big[\dfrac{1}{a}\ln(ax + b)\Big] = \dfrac{1}{a} \cdot \dfrac{a}{ax + b} = \dfrac{1}{ax + b} by the chain rule. For ax + b < 0, \ln\lvert ax + b \rvert = \ln(-ax - b), with derivative \dfrac{-a}{-ax - b} = \dfrac{a}{ax + b}, and the factor \dfrac{1}{a} again gives \dfrac{1}{ax + b}. On each interval where ax + b \ne 0, every antiderivative differs from this one by a constant.
The function \dfrac{1}{x - c} is defined for x < c, where x - c is negative and \ln(x - c) does not exist; \ln\lvert x - c \rvert covers both sides. But x^2 + c^2 \ge c^2 > 0 for every x, so \ln\left(x^2 + c^2\right) is already defined everywhere and equals \ln\left\lvert x^2 + c^2 \right\rvert.
k = -2, antiderivative \dfrac{x^2}{2} + C for x \ne 2. The remainder on dividing by x - 2 is the top’s value at 2, which is 4 + 2k, and it is 0 only for k = -2. Then \dfrac{x^2 - 2x}{x - 2} = x.
Division gives x^2 + 3x + 1 = (x + 1)(x + 2) - 1, so g(x) = x + 2 - \dfrac{1}{x + 1}. For x \ge 0 the gap \dfrac{1}{x + 1} is positive, so g lies below the line, and it tends to 0 as x \to \infty. The area between them is \int_0^b \dfrac{1}{x + 1}\,dx = \ln(b + 1), which grows without bound.