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Chapter 7 · Integrals

Exercise 7.4 · Q1: ∫3x2x6+1 dx\int\frac{3x^2}{x^6+1}\,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"):

3x2x6+1\frac{3x^2}{x^6 + 1}
Exam answer

Let I=∫3x2x6+1 dxI = \displaystyle\int\frac{3x^2}{x^6 + 1}\,dx⁠.

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

∴ I=∫dtt2+1=tan⁡−1t+C\therefore\ I = \int\frac{dt}{t^2 + 1} = \tan^{-1}t + C

∴ I=tan⁡−1(x3)+C\therefore\ I = \tan^{-1}(x^3) + C⁠.

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

The one idea this problem is built on

Read x6x^6 as (x3)2(x^3)^2⁠. Then the denominator is "something squared plus 1", and the numerator 3x23x^2 is exactly the derivative of that something, x3x^3⁠. The substitution t=x3t = x^3 therefore turns the integral into the standard shape ∫dtt2+1\int\frac{dt}{t^2 + 1}⁠, which is tan⁡−1t\tan^{-1} t by formula 7.4 (3).

Given, to find, and the plan
Given
f(x)=3x2x6+1f(x) = \dfrac{3x^2}{x^6 + 1}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Find where ff is defined. (2) Rewrite x6=(x3)2x^6 = (x^3)^2 and substitute t=x3t = x^3⁠. (3) Use formula 7.4 (3) with a=1a = 1⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Powers of powers

(xm)n=xmn(x^m)^n = x^{mn}⁠, so x6=(x3)2x^6 = (x^3)^2⁠. Basics: powers of x

2Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠; a factor g′(x) dxg'(x)\,dx in the integrand becomes dtdt⁠. Basics: substitution (the chain rule backwards)

3Formula 7.4 (3)

∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C⁠. With a=1a = 1 it is the standard formula ∫dx1+x2=tan⁡−1x+C\int\frac{dx}{1 + x^2} = \tan^{-1}x + C of Section 7.2. Lesson: integrals of particular functions

4Checking by differentiating

FF is an antiderivative of ff exactly when F′=fF' = f⁠. Basics: integration undoes differentiation

Solution

  1. Where is the integrand defined?

    Since x6=(x3)2≥0x^6 = (x^3)^2 \ge 0⁠, the denominator satisfies x6+1≥1x^6 + 1 \ge 1⁠. It is never 0, so f(x)f(x) is defined for every real xx⁠.

    An answer is only claimed where the integrand exists. Here there is no restriction at all, so the antiderivative we find will be valid on the whole real line.

  2. See the hidden square and the hidden derivative

    3x2x6+1=3x2(x3)2+1.\frac{3x^2}{x^6 + 1} = \frac{3x^2}{(x^3)^2 + 1}.

    The quantity being squared is x3x^3⁠, and its derivative is ddxx3=3x2\frac{d}{dx}x^3 = 3x^2⁠, which is exactly the numerator.

    Formula 7.4 (3) needs "(variable)2^2 + (constant)2^2" in the denominator and just "dd(variable)" on top. Writing x6x^6 as a square shows which variable to use, and the numerator confirms the choice: nothing is left over.
    (x3)2=x3⋅2=x6(x^3)^2 = x^{3\cdot 2} = x^6⁠. Basics: powers of x

  3. Substitute t=x3t = x^3

    Put t=x3t = x^3⁠. Then

    dtdx=3x2,so3x2 dx=dt.\frac{dt}{dx} = 3x^2, \quad\text{so}\quad 3x^2\,dx = dt.

    The integral becomes

    ∫3x2 dx(x3)2+1=∫dtt2+1.\int\frac{3x^2\,dx}{(x^3)^2 + 1} = \int\frac{dt}{t^2 + 1}.
    Every xx in the integrand disappears: (x3)2(x^3)^2 becomes t2t^2 and 3x2 dx3x^2\,dx becomes dtdt⁠. As xx runs over all reals, t=x3t = x^3 also runs over all reals, so no values are lost.
    Integration by substitution, Section 7.3.1. Lesson: integration by substitution

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

    ∫dtt2+12=11tan⁡−1t1+C=tan⁡−1t+C.\int\frac{dt}{t^2 + 1^2} = \frac11\tan^{-1}\frac t1 + C = \tan^{-1}t + C.
    The shape t2+a2t^2 + a^2 with a=1a = 1 is formula 7.4 (3). With a=1a = 1 the factor 1a\frac1a and the division inside tan⁡−1\tan^{-1} both disappear, leaving the basic formula (x) of Section 7.2.
    Formula 7.4 (3). Lesson: integrals of particular functions

  5. Return to xx

    ∫3x2x6+1 dx=tan⁡−1(x3)+C.\int\frac{3x^2}{x^6 + 1}\,dx = \tan^{-1}(x^3) + C.
    The question is about xx⁠, so the answer must be in xx⁠: replace tt by x3x^3⁠.

  6. Check: differentiate the answer back

    Let F(x)=tan⁡−1(x3)F(x) = \tan^{-1}(x^3)⁠. Chain rule, outer tan⁡−1u\tan^{-1}u⁠, inner u=x3u = x^3⁠:

    F′(x)=11+(x3)2⋅3x2=3x2x6+1.F'(x) = \frac{1}{1 + (x^3)^2}\cdot 3x^2 = \frac{3x^2}{x^6 + 1}.

    This is the integrand, for every real xx⁠. ✓

    ddutan⁡−1u=11+u2\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2}⁠, and the chain rule multiplies by the inner derivative 3x23x^2⁠. Getting back exactly the integrand is the definition of an antiderivative (Property (I), Section 7.2.1).
    Basics: integration undoes differentiation
Result

∫3x2x6+1 dx=tan⁡−1(x3)+C,\int\frac{3x^2}{x^6 + 1}\,dx = \tan^{-1}(x^3) + C, valid for every real xx⁠. (Textbook answer: tan⁡−1x3+C\tan^{-1}x^3 + C⁠, the same function.)

Check it with numbers

Take x=1x = 1 and h=0.001h = 0.001⁠:

  • F(1.001)=0.78689741F(1.001) = 0.78689741 and F(0.999)=0.78389742F(0.999) = 0.78389742⁠,
  • slope ≈0.78689741−0.783897420.002=1.5000\approx \frac{0.78689741 - 0.78389742}{0.002} = 1.5000⁠,
  • integrand: 3⋅1216+1=32=1.5000\frac{3\cdot 1^2}{1^6 + 1} = \frac32 = 1.5000⁠. ✓
Common mistakes
  1. Reading x6+1x^6 + 1 as x2+1x^2 + 1 with a power on top. The answer is not tan⁡−1x\tan^{-1}x or tan⁡−1(x6)\tan^{-1}(x^6)⁠; the variable inside tan⁡−1\tan^{-1} is the quantity that is squared, x3x^3⁠.
  2. Forgetting the differential. t=x3t = x^3 gives dt=3x2 dxdt = 3x^2\,dx⁠, not dt=dxdt = dx⁠. Here the 3 in the numerator is exactly what is needed; with a different numerator a constant factor would appear.
  3. Expecting a logarithm. The numerator 3x23x^2 is not the derivative of the denominator x6+1x^6 + 1 (that would be 6x56x^5⁠), so this is not a log⁡\log integral.
What to take away
  1. Look for "something squared plus a constant" in a denominator, with the derivative of that something on top: substitute and use 7.4 (3).
  2. Even powers hide squares: x4=(x2)2x^4 = (x^2)^2⁠, x6=(x3)2x^6 = (x^3)^2⁠.
  3. Always differentiate the answer back; the chain rule must reproduce the numerator.

Every fact used, and where it comes from

FactUsed inSource
(xm)n=xmn(x^m)^n = x^{mn}Step 2Laws of indices (Basics: powers of x)
Substitution t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dxStep 3Section 7.3.1 (Lesson: substitution)
∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + CStep 4Formula 7.4 (3), Section 7.4
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}Step 6Standard formula (x), Section 7.2
Chain rule; F′=fF' = f checks the answerStep 6Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
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