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Chapter 7 · Integrals

Exercise 7.4 · Q8: ∫x2x6+a6 dx\int\frac{x^2}{\sqrt{x^6+a^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"):

x2x6+a6\frac{x^2}{\sqrt{x^6 + a^6}}
Exam answer

Let I=∫x2x6+a6 dxI = \displaystyle\int\frac{x^2}{\sqrt{x^6 + a^6}}\,dx⁠.

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

I=13∫dtt2+(a3)2I = \frac13\int\frac{dt}{\sqrt{t^2 + \left(a^3\right)^2}}

[Using ∫dtt2+a2\displaystyle\int\frac{dt}{\sqrt{t^2 + a^2}} =log⁡∣t+t2+a2∣+C= \log\left|t + \sqrt{t^2 + a^2}\right| + C⁠, with a3a^3 in place of aa]

I=13log⁡∣t+t2+a6∣+CI = \frac13\log\left|t + \sqrt{t^2 + a^6}\right| + C

∴ I=13log⁡∣x3+x6+a6∣\therefore\ I = \dfrac13\log\left|x^3 + \sqrt{x^6 + a^6}\right| +C+ C (with a≠0a \ne 0⁠).

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

The one idea this problem is built on

Both sixth powers are squares of cubes: x6+a6=(x3)2+(a3)2x^6 + a^6 = (x^3)^2 + (a^3)^2⁠. The numerator x2 dxx^2\,dx is one third of d(x3)d(x^3)⁠. So t=x3t = x^3 gives 13∫dtt2+(a3)2\frac13\int\frac{dt}{\sqrt{t^2 + (a^3)^2}}⁠, which is formula 7.4 (6) with the constant a3a^3 playing the role of "aa".

Given, to find, and the plan
Given
f(x)=x2x6+a6f(x) = \dfrac{x^2}{\sqrt{x^6 + a^6}}⁠, where aa is a nonzero constant.
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 (6) with constant a3a^3⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Powers of powers

x6=(x3)2x^6 = (x^3)^2 and a6=(a3)2a^6 = (a^3)^2⁠. Basics: powers of x

2Substitution

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)

3Formula 7.4 (6)

∫dxx2+k2=log⁡∣x+x2+k2∣+C\int\frac{dx}{\sqrt{x^2 + k^2}} = \log\left|x + \sqrt{x^2 + k^2}\right| + C⁠, for any constant k≠0k \ne 0⁠. Lesson: integrals of particular functions

Solution

  1. Where is the integrand defined?

    With a≠0a \ne 0⁠, a6>0a^6 > 0 and x6≥0x^6 \ge 0⁠, so x6+a6>0x^6 + a^6 > 0 for every real xx⁠. The integrand is defined on the whole real line.

    The book does not say so, but a≠0a \ne 0 is needed: formula 7.4 (6) is proved with x=atan⁡θx = a\tan\theta⁠, which needs a nonzero constant. If a=0a = 0 the integrand would be x2∣x∣3=1∣x∣\frac{x^2}{|x|^3} = \frac{1}{|x|}⁠, undefined at 0, and the answer below would break down for x<0x < 0⁠, where x3+x6=x3+∣x3∣=0x^3 + \sqrt{x^6} = x^3 + |x^3| = 0⁠. We take a≠0a \ne 0 throughout.

  2. Substitute t=x3t = x^3

    Put t=x3t = x^3⁠. Then dt=3x2 dxdt = 3x^2\,dx⁠, x2 dx=13 dtx^2\,dx = \frac13\,dt⁠, and x6+a6=t2+(a3)2x^6 + a^6 = t^2 + (a^3)^2⁠:

    ∫x2 dxx6+a6=13∫dtt2+(a3)2.\int\frac{x^2\,dx}{\sqrt{x^6 + a^6}} = \frac13\int\frac{dt}{\sqrt{t^2 + (a^3)^2}}.
    The numerator x2x^2 is a third of the derivative of x3x^3⁠, and the root contains (x3)2(x^3)^2⁠: both point to t=x3t = x^3⁠. As xx runs over all reals, so does tt⁠.
    Integration by substitution, Section 7.3.1. Lesson: integration by substitution

  3. Use formula 7.4 (6) with the constant a3a^3

    13∫dtt2+(a3)2=13log⁡∣t+t2+a6∣+C.\frac13\int\frac{dt}{\sqrt{t^2 + (a^3)^2}} = \frac13\log\left|t + \sqrt{t^2 + a^6}\right| + C.
    Formula 7.4 (6) only involves the square of its constant, so it does not matter that a3a^3 is negative when aa is negative: (a3)2=a6(a^3)^2 = a^6 either way.
    Formula 7.4 (6). Lesson: integrals of particular functions

  4. Return to xx

    ∫x2x6+a6 dx=13log⁡∣x3+x6+a6∣+C.\int\frac{x^2}{\sqrt{x^6 + a^6}}\,dx = \frac13\log\left|x^3 + \sqrt{x^6 + a^6}\right| + C.

    The quantity inside the modulus is positive for every xx⁠, since x6+a6>x6=∣x3∣\sqrt{x^6 + a^6} > \sqrt{x^6} = |x^3|⁠.

    Replace tt by x3x^3 and t2t^2 by x6x^6⁠. The book keeps the modulus, which is harmless.

  5. Check: differentiate the answer back

    Let s=x6+a6s = \sqrt{x^6 + a^6}⁠, so s′=6x52s=3x5ss' = \frac{6x^5}{2s} = \frac{3x^5}{s}⁠. Then, with F(x)=13log⁡(x3+s)F(x) = \frac13\log(x^3 + s)⁠,

    F′(x)=13⋅3x2+3x5sx3+s=13⋅3x2(s+x3)s (x3+s)=x2s=x2x6+a6. ✓\begin{aligned} F'(x) &= \frac13\cdot\frac{3x^2 + \frac{3x^5}{s}}{x^3 + s} \\ &= \frac13\cdot\frac{3x^2(s + x^3)}{s\,(x^3 + s)} \\ &= \frac{x^2}{s} = \frac{x^2}{\sqrt{x^6 + a^6}}. \ \checkmark \end{aligned}
    Chain rule for log⁡u\log u with u=x3+su = x^3 + s⁠, and for v\sqrt v with v=x6+a6v = x^6 + a^6 (aa is a constant, so ddxa6=0\frac{d}{dx}a^6 = 0⁠). Factor 3x23x^2 out of the numerator: 3x2+3x5s=3x2(s+x3)s3x^2 + \frac{3x^5}{s} = \frac{3x^2(s + x^3)}{s}⁠, and s+x3s + x^3 cancels.
    Basics: integration undoes differentiation
Result

∫x2x6+a6 dx=13log⁡∣x3+x6+a6∣+C,\int\frac{x^2}{\sqrt{x^6 + a^6}}\,dx = \frac13\log\left|x^3 + \sqrt{x^6 + a^6}\right| + C, valid for every real xx (with a≠0a \ne 0⁠). (Textbook answer: 13log⁡∣x3+x6+a6∣+C\frac13\log\left|x^3 + \sqrt{x^6 + a^6}\right| + C⁠, the same.)

Check it with numbers

Take a=1a = 1⁠, x=1x = 1 and h=0.001h = 0.001⁠:

  • F(1.001)=0.29449848F(1.001) = 0.29449848 and F(0.999)=0.29308427F(0.999) = 0.29308427⁠,
  • slope ≈0.29449848−0.293084270.002=0.7071\approx \frac{0.29449848 - 0.29308427}{0.002} = 0.7071⁠,
  • integrand: 11+1=12=0.7071\frac{1}{\sqrt{1 + 1}} = \frac{1}{\sqrt2} = 0.7071⁠. ✓
Common mistakes
  1. Taking the formula's constant to be aa or a6a^6⁠. It is the number whose square is a6a^6⁠, namely a3a^3 (or ∣a∣3|a|^3⁠).
  2. Losing the 13\frac13 from x2 dx=13 dtx^2\,dx = \frac13\,dt⁠.
  3. Answering with sin⁡−1\sin^{-1} or tan⁡−1\tan^{-1}⁠. A root of "square plus constant" gives a logarithm.
What to take away
  1. Sixth powers are squares of cubes; with x2x^2 on top, try t=x3t = x^3⁠.
  2. The "aa" in a formula can be any nonzero constant expression, here a3a^3⁠.
  3. State the assumptions a question leaves unsaid (here a≠0a \ne 0⁠).

Every fact used, and where it comes from

FactUsed inSource
(x3)2=x6(x^3)^2 = x^6⁠, (a3)2=a6(a^3)^2 = a^6Step 2Laws of indices (Basics: powers of x)
Substitution t=x3t = x^3⁠, dt=3x2 dxdt = 3x^2\,dxStep 2Section 7.3.1 (Lesson: substitution)
∫dxx2+a2=log⁡∣x+x2+a2∣+C\int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + CStep 3Formula 7.4 (6), Section 7.4
Chain rule; F′=fF' = f checks the answerStep 5Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
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