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Chapter 7 · Integrals

Exercise 7.1 · Q13: ∫x3−x2+x−1x−1 dx\int \frac{x^3-x^2+x-1}{x-1}\,dx

Exercise 7.1Integration as anti-differentiation · Section 7.2
The question

Find the following integral:

∫x3−x2+x−1x−1 dx\int \frac{x^3 - x^2 + x - 1}{x - 1}\,dx
Exam answer

Let I=∫x3−x2+x−1x−1 dxI = \displaystyle\int\frac{x^3 - x^2 + x - 1}{x - 1}\,dx⁠.

x3−x2+x−1=x2(x−1)+(x−1)=(x−1)(x2+1)\begin{aligned} &x^3 - x^2 + x - 1 \\ &= x^2(x - 1) + (x - 1) \\ &= (x - 1)\left(x^2 + 1\right) \end{aligned}

Cancelling x−1x - 1 (as x≠1x \ne 1⁠),

I=∫(x2+1)dxI = \int\left(x^2 + 1\right)dx

∴ I=x33+x+C\therefore\ I = \dfrac{x^3}{3} + x + C⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The denominator x−1x - 1 has two terms, so we cannot split the fraction term by term. Instead we factor the numerator: grouping gives x3−x2+x−1=(x−1)(x2+1)x^3 - x^2 + x - 1 = (x-1)(x^2+1)⁠. The factor x−1x - 1 cancels (wherever x≠1x \ne 1⁠), and what is left, x2+1x^2 + 1⁠, is a plain polynomial that the power formula integrates at once.

Given, to find, and the plan
Given
The integrand f(x)=x3−x2+x−1x−1f(x) = \dfrac{x^3 - x^2 + x - 1}{x - 1}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠.
Plan
(1) Note where ff is defined. (2) Factor the numerator by grouping and cancel x−1x - 1⁠. (3) Integrate the polynomial term by term. (4) Differentiate the answer to check it.

Basics you need

1Integration undoes differentiation

∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C means F′(x)=f(x)F'(x) = f(x)⁠. Basics: integration undoes differentiation

2Factoring by grouping

Pair the terms so that each pair shares a factor, take it out, and look for a common bracket: x3−x2=x2(x−1)x^3 - x^2 = x^2(x - 1) and x−1=1⋅(x−1)x - 1 = 1\cdot(x - 1)⁠. Basics: splitting and simplifying fractions

3Cancelling a common factor

(x−1) g(x)x−1=g(x)\dfrac{(x-1)\,g(x)}{x-1} = g(x) is true only when x−1≠0x - 1 \ne 0⁠: we may never divide by zero. Basics: splitting fractions

4The power formula

∫xn dx=xn+1n+1+C\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1 (formula (i), Section 7.2); ∫dx=x+C\int dx = x + C⁠. Basics: powers of x and the power rule

Solution

  1. Where is the integrand defined?

    The denominator x−1x - 1 is zero at x=1x = 1⁠, so the integrand is defined for every real xx except 1:

    x≠1.x \ne 1.
    We need this before cancelling anything. It also tells us the answer will be valid on the two intervals (−∞,1)(-\infty, 1) and (1,∞)(1, \infty)⁠.
    The Note in Section 7.2: the interval on which a function is defined must be kept in mind in every problem.

  2. Factor the numerator by grouping

    x3−x2+x−1=(x3−x2)+(x−1)=x2(x−1)+1⋅(x−1)=(x−1)(x2+1).\begin{aligned} x^3 - x^2 + x - 1 &= (x^3 - x^2) + (x - 1) \\ &= x^2(x - 1) + 1\cdot(x - 1) \\ &= (x - 1)(x^2 + 1). \end{aligned}
    We look for the factor x−1x - 1 because it is the denominator: if it divides the numerator, the fraction collapses to a polynomial. The first pair has the common factor x2x^2⁠; after taking it out, both pairs contain (x−1)(x - 1)⁠, which then comes out of the whole sum. Expanding (x−1)(x2+1)=x3+x−x2−1(x-1)(x^2+1) = x^3 + x - x^2 - 1 confirms the factorisation.
    Another way to see that x−1x - 1 is a factor: the numerator is 0 at x=1x = 1 (1−1+1−1=01 - 1 + 1 - 1 = 0⁠), so by the factor theorem x−1x - 1 divides it. Long division gives the same quotient x2+1x^2 + 1 with remainder 0.

  3. Cancel x−1x - 1

    For x≠1x \ne 1⁠,

    x3−x2+x−1x−1=(x−1)(x2+1)x−1=x2+1.\frac{x^3 - x^2 + x - 1}{x - 1} = \frac{(x-1)(x^2+1)}{x-1} = x^2 + 1.
    Cancelling is division of top and bottom by x−1x - 1⁠, which is allowed precisely because x−1≠0x - 1 \ne 0 on the domain found in Step 1. So on its domain the integrand is the function x2+1x^2 + 1⁠; the only difference is a single missing point at x=1x = 1 (the figure below).
    Simplifying the integrand before integrating is the main tool of this exercise. Basics: splitting fractions

  4. Integrate term by term

    ∫(x2+1) dx=∫x2 dx+∫1 dx=x33+x+C.\begin{aligned} \int (x^2 + 1)\,dx &= \int x^2\,dx + \int 1\,dx \\ &= \frac{x^3}{3} + x + C. \end{aligned}
    Property III lets us integrate a sum one term at a time. Formula (i) with n=2n = 2 gives x33\frac{x^{3}}{3}⁠; with n=0n = 0 it gives ∫dx=x\int dx = x⁠. The constants of the two parts are combined into one CC⁠.
    Formula (i), Section 7.2, and its special case ∫dx=x+C\int dx = x + C⁠. Basics: the power rule

  5. Check: differentiate the answer back

    ddx(x33+x+C)=3x23+1+0=x2+1=x3−x2+x−1x−1(x≠1).\begin{aligned} \frac{d}{dx}\left(\frac{x^3}{3} + x + C\right) &= \frac{3x^2}{3} + 1 + 0 \\ &= x^2 + 1 \\ &= \frac{x^3 - x^2 + x - 1}{x - 1}\quad (x \ne 1). \end{aligned}

    The derivative is the integrand at every point of its domain, so the answer is correct.

    The last equality is Step 3 read backwards, and it needs x≠1x \ne 1⁠. So the answer is an antiderivative on (−∞,1)(-\infty, 1) and on (1,∞)(1, \infty)⁠. Strictly, since the domain has two pieces, the constant could even be chosen differently on each piece; like the textbook, we write a single CC⁠.
    Property I, Section 7.2.1: differentiation undoes integration. Basics: integration undoes differentiation
x y −2 −1 1 2 2 4 6 8 hole at (1, 2) y = x² + 1, x ≠ 1
The integrand equals x2+1x^2 + 1 at every point except x=1x = 1⁠, where the original fraction is 00\frac00 and undefined: the graph is the parabola with one hole. Integrating it is the same as integrating x2+1x^2 + 1 on each side of the hole.
Result

For x≠1x \ne 1⁠,

∫x3−x2+x−1x−1 dx=x33+x+C.\int \frac{x^3 - x^2 + x - 1}{x - 1}\,dx = \frac{x^3}{3} + x + C.

(Textbook answer: x33+x+C\frac{x^3}{3} + x + C⁠.)

Check it with numbers

At x=3x = 3 the integrand is 27−9+3−12=202=10\frac{27 - 9 + 3 - 1}{2} = \frac{20}{2} = 10⁠, and 32+1=103^2 + 1 = 10 too, as Step 3 says. With F(x)=x33+xF(x) = \frac{x^3}{3} + x⁠: F(3.001)≈12.01000300F(3.001) \approx 12.01000300 and F(2.999)≈11.99000300F(2.999) \approx 11.99000300⁠, so the slope of FF at 3 is about

12.01000300−11.990003000.002≈10.0000.\frac{12.01000300 - 11.99000300}{0.002} \approx 10.0000.

It matches the integrand value 10.

(Each value of FF is rounded to 8 decimal places, so the slope is reliable to about 4 decimal places.)

Common mistakes
  1. Splitting over a two-term denominator. x3x−1\frac{x^3}{x - 1} is not x3x−x31\frac{x^3}{x} - \frac{x^3}{1}⁠. Only a single-term denominator can be divided into each term.
  2. Grouping signs wrongly. −x2+x−1-x^2 + x - 1 is not −(x2+x−1)-(x^2 + x - 1)⁠. Group as (x3−x2)+(x−1)(x^3 - x^2) + (x - 1)⁠.
  3. Forgetting x≠1x \ne 1⁠. The cancellation, and so the answer, holds only away from x=1x = 1⁠.
  4. Writing ∫1 dx=0\int 1\,dx = 0 or 11⁠. The integral of the constant 1 is x+Cx + C⁠.
What to take away
  1. When the denominator has several terms, try to factor the numerator and cancel; test a candidate factor x−ax - a by substituting x=ax = a⁠.
  2. A cancelled factor leaves a hole in the domain; state it with the answer.
  3. Differentiate back to check: ddx(x33+x)=x2+1\frac{d}{dx}\left(\frac{x^3}{3} + x\right) = x^2 + 1⁠.

Every fact used, and where it comes from

FactUsed inSource
Keep the interval of definition in mindSteps 1, 5Note after the list of formulae, Section 7.2
x3−x2+x−1=(x−1)(x2+1)x^3 - x^2 + x - 1 = (x-1)(x^2+1)Step 2Factoring by grouping; factor theorem (earlier classes)
(x−1)gx−1=g\frac{(x-1)g}{x-1} = g for x≠1x \ne 1Step 3Arithmetic of fractions (no division by zero)
∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C⁠; ∫dx=x+C\int dx = x + CStep 4Formula (i), Section 7.2
∫(f+g) dx=∫f dx+∫g dx\int (f + g)\,dx = \int f\,dx + \int g\,dxStep 4Property III, Section 7.2.1 (Lesson: integration as anti-differentiation)
ddx∫f dx=f\frac{d}{dx}\int f\,dx = fStep 5Property I, Section 7.2.1
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