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Chapter 7 · Integrals

Exercise 7.1 · Q7: ∫x2(1−1x2)dx\int x^2\left(1 - \frac{1}{x^2}\right)dx

Exercise 7.1Properties of indefinite integrals · Section 7.2.1
The question

Find the following integral:

∫x2(1−1x2)dx\int x^2\left(1 - \frac{1}{x^2}\right)dx
Exam answer

Let I=∫x2(1−1x2)dxI = \displaystyle\int x^2\left(1 - \frac{1}{x^2}\right)dx⁠.

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

There is no "product rule" for integrals, so we do not integrate the two factors separately. Instead we simplify first: multiplying out gives x2−1x^2 - 1⁠, a polynomial, and polynomials are integrated term by term with the power rule.

Given, to find, and the plan
Given
The integrand x2(1−1x2)x^2\left(1 - \frac{1}{x^2}\right)⁠, defined for every real x≠0x \ne 0 (because of 1x2\frac{1}{x^2}⁠).
To find
The indefinite integral.
Plan
(1) Note the domain. (2) Multiply out and simplify. (3) Integrate term by term. (4) Differentiate back to check. (5) Say where the answer is valid.

Basics you need

1Power rule for integrals

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

2Simplify before integrating

Multiplying out a bracket or splitting a fraction often turns an awkward integrand into a sum of powers. Basics: splitting fractions

3Term by term

∫(f−g) dx=∫f dx−∫g dx\int (f - g)\,dx = \int f\,dx - \int g\,dx (Properties (III) to (V), Section 7.2.1). Lesson: integration as anti differentiation

Solution

  1. Note where the integrand is defined

    The term 1x2\frac{1}{x^2} needs x≠0x \ne 0⁠. So the integrand is defined on the two intervals x<0x < 0 and x>0x > 0⁠, and that is where we will look for anti derivatives.

    An anti derivative is defined on an interval where the integrand exists. Simplifying will hide the 1x2\frac{1}{x^2}⁠, so we record the restriction now, before it disappears from view.
    Note after the standard formulae, Section 7.2: in any specific problem we must keep the interval of definition in mind. Basics: integration undoes differentiation

  2. Multiply out

    For x≠0x \ne 0⁠, distribute x2x^2 over the bracket:

    x2(1−1x2)=x2⋅1−x2⋅1x2=x2−1.\begin{aligned} x^2\left(1 - \frac{1}{x^2}\right) &= x^2\cdot 1 - x^2\cdot\frac{1}{x^2} \\ &= x^2 - 1. \end{aligned}
    x2⋅1x2=1x^2\cdot\frac{1}{x^2} = 1 because x2≠0x^2 \ne 0⁠. The integrand is now a polynomial with two terms, each a standard power.
    Distributive law a(b−c)=ab−aca(b - c) = ab - ac⁠. Basics: splitting fractions

  3. Integrate term by term (power rule)

    ∫(x2−1)dx=∫x2 dx−∫1 dx=x2+12+1−x+C=x33−x+C.\begin{aligned} \int\left(x^2 - 1\right)dx &= \int x^2\,dx - \int 1\,dx \\ &= \frac{x^{2+1}}{2+1} - x + C \\ &= \frac{x^3}{3} - x + C. \end{aligned}
    Property (III) splits the difference into two integrals. For x2x^2 the power rule with n=2n = 2 raises the power to 33 and divides by 33⁠. The constant 11 integrates to xx⁠. The two constants of integration are combined into one CC⁠.
    Standard formulae, item (i), Section 7.2; Property (V), Section 7.2.1. Basics: the power rule

  4. Check: differentiate the answer back

    ddx(x33−x+C)=3x23−1=x2−1=x2(1−1x2), x≠0.  ✓\begin{aligned} \frac{d}{dx}\left(\frac{x^3}{3} - x + C\right) &= \frac{3x^2}{3} - 1 \\ &= x^2 - 1 \\ &= x^2\left(1 - \frac{1}{x^2}\right),\ x \ne 0. \;\checkmark \end{aligned}
    The derivative returns the simplified integrand, and Step 2 showed that equals the original integrand whenever x≠0x \ne 0⁠.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation

  5. Where the answer is valid

    The formula x33−x+C\frac{x^3}{3} - x + C gives the anti derivatives on any interval that does not contain 00 (for instance on x>0x > 0⁠, or on x<0x < 0⁠).

    The polynomial x33−x\frac{x^3}{3} - x itself makes sense at x=0x = 0⁠, but the given integrand does not, so the question only concerns intervals avoiding 00⁠. The book's "Note" says we normally do not mention the interval, but we must keep it in mind.
    Note after the standard formulae, Section 7.2. Lesson: integration as anti differentiation
Result ∫x2(1−1x2)dx=x33−x+C,\int x^2\left(1 - \frac{1}{x^2}\right)dx = \frac{x^3}{3} - x + C,

on any interval not containing 00⁠. (Textbook answer: x33−x+C\frac{x^3}{3} - x + C⁠.)

Check it with numbers

At x=2x = 2 with h=0.001h = 0.001⁠, for F(x)=x33−xF(x) = \frac{x^3}{3} - x⁠:

F(2.001)−F(1.999)0.002=0.669668667000−0.6636686663330.002≈3.0000003.\begin{aligned} &\frac{F(2.001) - F(1.999)}{0.002} \\ &= \frac{0.669668667000 - 0.663668666333}{0.002} \\ &\approx 3.0000003. \end{aligned}

The integrand at x=2x = 2 is 4(1−14)=34\left(1 - \frac14\right) = 3⁠. ✓

Common mistakes
  1. Integrating factor by factor. ∫x2 dx\int x^2\,dx multiplied by ∫(1−1x2)dx\int\left(1 - \frac{1}{x^2}\right)dx is not the integral of the product; there is no such rule.
  2. x2⋅1x2=xx^2\cdot\frac{1}{x^2} = x or 00⁠. It equals 11⁠, so the integrand is x2−1x^2 - 1⁠, not x2−xx^2 - x or x2x^2⁠.
  3. Writing −1-1 instead of −x-x⁠. The constant 11 integrates to xx⁠, so the second term of the answer is −x-x⁠.
What to take away
  1. Simplify the integrand into a sum of powers before integrating.
  2. Note the domain of the original integrand before simplifying hides it.

Every fact used, and where it comes from

FactUsed inSource
Keep the interval of definition in mindSteps 1, 5Note after the standard formulae, Section 7.2
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1⁠; ∫dx=x+C\int dx = x + CStep 3Standard formulae, item (i), Section 7.2
Integrate term by termStep 3Properties (III) to (V), Section 7.2.1
Differentiation and integration undo each otherStep 4Property (I), Section 7.2.1
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