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Chapter 7 · Integrals

Exercise 7.4 · Q6: ∫x21−x6 dx\int\frac{x^2}{1-x^6}\,dx

Exercise 7.4Integrals of some particular functions · Section 7.4
The question

Integrate the function (Exercise 7.4, "Integrate the functions in Exercises 1 to 23"):

x21−x6\frac{x^2}{1 - x^6}
Exam answer

Let I=∫x21−x6 dxI = \displaystyle\int\frac{x^2}{1 - x^6}\,dx⁠.

Put x3=tx^3 = t⁠. Then 3x2 dx=dt3x^2\,dx = dt⁠.

I=13∫dt1−t2I = \frac13\int\frac{dt}{1 - t^2}

Using ∫dta2−t2\displaystyle\int\frac{dt}{a^2 - t^2} =12alog⁡∣a+ta−t∣+C= \frac{1}{2a}\log\left|\frac{a + t}{a - t}\right| + C⁠,

I=13⋅12log⁡∣1+t1−t∣+CI = \frac13\cdot\frac12\log\left|\frac{1 + t}{1 - t}\right| + C

∴ I=16log⁡∣1+x31−x3∣+C\therefore\ I = \dfrac16\log\left|\dfrac{1 + x^3}{1 - x^3}\right| + C⁠.

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

The one idea this problem is built on

x6=(x3)2x^6 = (x^3)^2⁠, and x2 dxx^2\,dx is one third of d(x3)d(x^3)⁠. So t=x3t = x^3 turns the integral into 13∫dt1−t2\frac13\int\frac{dt}{1 - t^2}⁠, which is formula 7.4 (2): ∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + C⁠. No square root here, so the answer is a logarithm, not sin⁡−1\sin^{-1}⁠.

Given, to find, and the plan
Given
f(x)=x21−x6f(x) = \dfrac{x^2}{1 - x^6}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Find where ff is defined. (2) Substitute t=x3t = x^3⁠. (3) Use formula 7.4 (2) with a=1a = 1⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Substitution

t=x3t = x^3 gives dt=3x2 dxdt = 3x^2\,dx⁠, so x2 dx=13 dtx^2\,dx = \frac13\,dt⁠. Basics: substitution (the chain rule backwards)

2Formula 7.4 (2)

∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + C⁠. It comes from splitting 1a2−x2\frac{1}{a^2 - x^2} into two simple fractions. Basics: splitting fractions

3∫u′u dx=log⁡∣u∣+C\int\frac{u'}{u}\,dx = \log|u| + C

Used in the check. Basics: logarithms and 1/x

Solution

  1. Where is the integrand defined?

    The denominator vanishes when x6=1x^6 = 1⁠, that is, x=1x = 1 or x=−1x = -1⁠. So ff is defined for x≠±1x \ne \pm 1⁠: on the three intervals x<−1x < -1⁠, −1<x<1-1 < x < 1 and x>1x > 1⁠.

    The real solutions of x6=1x^6 = 1 are only ±1\pm 1 (an even power equals 1 exactly when ∣x∣=1|x| = 1⁠). The answer will be valid on each of the three intervals separately, with its own constant on each.

  2. Substitute t=x3t = x^3

    Put t=x3t = x^3⁠. Then dt=3x2 dxdt = 3x^2\,dx⁠, so x2 dx=13 dtx^2\,dx = \frac13\,dt⁠, and x6=t2x^6 = t^2⁠:

    ∫x2 dx1−x6=13∫dt1−t2.\int\frac{x^2\,dx}{1 - x^6} = \frac13\int\frac{dt}{1 - t^2}.

    The excluded points x=±1x = \pm 1 correspond to t=±1t = \pm 1⁠, where the new integrand is also undefined.

    The numerator x2x^2 is (one third of) the derivative of x3x^3⁠, and x6x^6 is the square of x3x^3⁠. Both point to the same substitution.
    Integration by substitution, Section 7.3.1. Lesson: integration by substitution

  3. Use formula 7.4 (2) with a=1a = 1

    13∫dt12−t2=13⋅12log⁡∣1+t1−t∣+C.\frac13\int\frac{dt}{1^2 - t^2} = \frac13\cdot\frac{1}{2}\log\left|\frac{1 + t}{1 - t}\right| + C.
    "Constant squared minus a square" with no root is formula 7.4 (2). The book proves it by writing 1a2−t2=12a(1a−t+1a+t)\frac{1}{a^2 - t^2} = \frac{1}{2a}\left(\frac{1}{a - t} + \frac{1}{a + t}\right) and integrating each part to a logarithm; that works for every t≠±at \ne \pm a⁠.
    Formula 7.4 (2). Lesson: integrals of particular functions

  4. Return to xx

    ∫x21−x6 dx=16log⁡∣1+x31−x3∣+C.\int\frac{x^2}{1 - x^6}\,dx = \frac16\log\left|\frac{1 + x^3}{1 - x^3}\right| + C.
    13⋅12=16\frac13\cdot\frac12 = \frac16⁠, and tt is replaced by x3x^3⁠. The modulus matters here: for ∣x∣>1|x| > 1 the fraction 1+x31−x3\frac{1 + x^3}{1 - x^3} is negative.

  5. Check: differentiate the answer back

    Write F(x)=16(log⁡∣1+x3∣−log⁡∣1−x3∣)F(x) = \frac16\left(\log|1 + x^3| - \log|1 - x^3|\right)⁠. Using ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u}⁠:

    F′(x)=16(3x21+x3+3x21−x3)=x22⋅(1−x3)+(1+x3)(1+x3)(1−x3)=x22⋅21−x6=x21−x6. ✓\begin{aligned} F'(x) &= \frac16\left(\frac{3x^2}{1 + x^3} + \frac{3x^2}{1 - x^3}\right) \\ &= \frac{x^2}{2}\cdot\frac{(1 - x^3) + (1 + x^3)}{(1 + x^3)(1 - x^3)} \\ &= \frac{x^2}{2}\cdot\frac{2}{1 - x^6} = \frac{x^2}{1 - x^6}. \ \checkmark \end{aligned}
    The derivative of 1−x31 - x^3 is −3x2-3x^2⁠, so −log⁡∣1−x3∣-\log|1 - x^3| contributes +3x21−x3+\frac{3x^2}{1 - x^3}⁠. Over a common denominator the x3x^3 terms cancel, and (1+x3)(1−x3)=1−x6(1 + x^3)(1 - x^3) = 1 - x^6⁠. This holds for every x≠±1x \ne \pm1⁠.
    Standard formula (xii), Section 7.2, with the chain rule. Basics: integration undoes differentiation
Result

∫x21−x6 dx=16log⁡∣1+x31−x3∣+C,\int\frac{x^2}{1 - x^6}\,dx = \frac16\log\left|\frac{1 + x^3}{1 - x^3}\right| + C, valid for x≠±1x \ne \pm 1 (on each of the three intervals). (Textbook answer: 16log⁡∣1+x31−x3∣+C\frac16\log\left|\frac{1 + x^3}{1 - x^3}\right| + C⁠, the same.)

Check it with numbers

Take x=0.5x = 0.5 and h=0.001h = 0.001⁠:

  • F(0.501)=0.04214024F(0.501) = 0.04214024 and F(0.499)=0.04163230F(0.499) = 0.04163230⁠,
  • slope ≈0.04214024−0.041632300.002=0.2540\approx \frac{0.04214024 - 0.04163230}{0.002} = 0.2540⁠,
  • integrand: 0.251−164=1663=0.2540\frac{0.25}{1 - \frac{1}{64}} = \frac{16}{63} = 0.2540⁠. ✓
Common mistakes
  1. Answering with sin⁡−1\sin^{-1} or tan⁡−1\tan^{-1}⁠. sin⁡−1\sin^{-1} needs 1−t2\sqrt{1 - t^2} in the denominator and tan⁡−1\tan^{-1} needs 1+t21 + t^2⁠. Here it is 1−t21 - t^2 with no root: a logarithm.
  2. Using formula 7.4 (1) by mistake. That one is for t2−a2t^2 - a^2 and gives log⁡∣t−at+a∣\log\left|\frac{t - a}{t + a}\right|⁠, which is the negative of our answer; mixing them flips the sign.
  3. Dropping the modulus. For ∣x∣>1|x| > 1 the fraction is negative, and the log of a negative number is undefined.
  4. Writing 13\frac13 instead of 16\frac16⁠. The 13\frac13 from dtdt and the 12a=12\frac{1}{2a} = \frac12 both apply.
What to take away
  1. ∫dta2−t2=12alog⁡∣a+ta−t∣+C\int\frac{dt}{a^2 - t^2} = \frac{1}{2a}\log\left|\frac{a + t}{a - t}\right| + C⁠.
  2. Root or no root changes everything: 1a2−t2\frac{1}{\sqrt{a^2 - t^2}} gives sin⁡−1\sin^{-1}⁠, 1a2−t2\frac{1}{a^2 - t^2} gives a logarithm.
  3. Look for xn−1x^{n-1} on top and x2nx^{2n} below: t=xnt = x^n⁠.

Every fact used, and where it comes from

FactUsed inSource
Substitution t=x3t = x^3⁠, dt=3x2 dxdt = 3x^2\,dxStep 2Section 7.3.1 (Lesson: substitution)
∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + CStep 3Formula 7.4 (2), Section 7.4
ddxlog⁡∣x∣=1x\frac{d}{dx}\log|x| = \frac1x⁠; chain ruleStep 5Standard formula (xii), Section 7.2; Theorem 4, Section 5.3.1
(1+x3)(1−x3)=1−x6(1 + x^3)(1 - x^3) = 1 - x^6Step 5Difference of squares
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