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Chapter 7 · Integrals

Exercise 7.2 · Q12: ∫(x3−1)1/3x5 dx\int (x^3-1)^{1/3}x^5\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function:

(x3−1)13x5\left(x^3 - 1\right)^{\frac13}x^5
Exam answer

Let I=∫(x3−1)1/3x5 dxI = \displaystyle\int\left(x^3 - 1\right)^{1/3}x^5\,dx⁠.

Put x3−1=tx^3 - 1 = t⁠. Then 3x2 dx=dt3x^2\,dx = dt and x3=t+1x^3 = t + 1⁠.

I=∫(x3−1)1/3x3⋅x2 dx=13∫t1/3(t+1) dt=13∫(t4/3+t1/3)dt=13(t7/37/3+t4/34/3)+C=17 t7/3+14 t4/3+C\begin{aligned} I &= \int\left(x^3 - 1\right)^{1/3}x^3\cdot x^2\,dx \\ &= \frac13\int t^{1/3}(t + 1)\,dt \\ &= \frac13\int\left(t^{4/3} + t^{1/3}\right)dt \\ &= \frac13\left(\frac{t^{7/3}}{7/3} + \frac{t^{4/3}}{4/3}\right) + C \\ &= \frac17\,t^{7/3} + \frac14\,t^{4/3} + C \end{aligned}

∴ I=17(x3−1)7/3\therefore\ I = \dfrac17\left(x^3 - 1\right)^{7/3} +14(x3−1)4/3+C+ \dfrac14\left(x^3 - 1\right)^{4/3} + C⁠.

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

The one idea this problem is built on

The inside x3−1x^3 - 1 has derivative 3x23x^2⁠. Split x5=x3⋅x2x^5 = x^3\cdot x^2⁠: the x2x^2 goes with dxdx to make dt3\frac{dt}{3}⁠, and the leftover x3x^3 is rewritten as t+1t + 1⁠. Then the integrand is (t+1)t1/3(t+1)t^{1/3}⁠, a sum of two powers.

Given, to find, and the plan
Given
f(x)=(x3−1)1/3x5f(x) = (x^3-1)^{1/3}x^5⁠. Here u1/3u^{1/3} is the real cube root, defined for every real uu (for example (−8)1/3=−2(-8)^{1/3} = -2⁠). So ff is defined and continuous for every real xx⁠.
To find
∫(x3−1)1/3x5 dx\displaystyle\int(x^3-1)^{1/3}x^5\,dx⁠.
Plan
(1) Split x5=x3⋅x2x^5 = x^3\cdot x^2⁠. (2) Substitute t=x3−1t = x^3-1⁠, with x3=t+1x^3 = t+1 and x2 dx=dt3x^2\,dx = \frac{dt}{3}⁠. (3) Expand into powers. (4) Integrate. (5) Return to xx⁠. (6) Differentiate to check.

Basics you need

1Splitting a power

x5=x3⋅x2x^5 = x^3\cdot x^2⁠, and t⋅t1/3=t4/3t\cdot t^{1/3} = t^{4/3}⁠. Basics: powers of x

2Substitution, including the leftover xx

Put t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dx⁠, and rewrite any other xx through tt (Section 7.3.1). Basics: substitution

3The power formula

∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C⁠, n≠−1n \ne -1 (standard formula (i), Section 7.2).

Solution

  1. Split x5x^5 to expose the inside's derivative

    ddx(x3−1)=3x2,x5 dx=x3⋅(x2 dx).\begin{aligned} \frac{d}{dx}\left(x^3 - 1\right) &= 3x^2, \\ x^5\,dx &= x^3\cdot\left(x^2\,dx\right). \end{aligned}
    The derivative of the inside is 3x23x^2⁠, so we want an x2 dxx^2\,dx to turn into dtdt⁠. The integrand has x5x^5⁠, which is more than x2x^2⁠; the spare factor x3x^3 will be expressed through tt⁠.

  2. Substitute t=x3−1t = x^3 - 1

    t=x3−1,x3=t+1,dt=3x2 dx,x2 dx=dt3.\begin{aligned} t &= x^3 - 1, \\ x^3 &= t + 1, \\ dt &= 3x^2\,dx, \\ x^2\,dx &= \frac{dt}{3}. \end{aligned}

    Therefore

    ∫(x3−1)1/3 x3 (x2 dx)=∫t1/3(t+1) dt3=13∫(t+1) t1/3 dt.\begin{aligned} \int (x^3-1)^{1/3}\,x^3\,(x^2\,dx) &= \int t^{1/3}(t+1)\,\frac{dt}{3} \\ &= \frac13\int (t+1)\,t^{1/3}\,dt. \end{aligned}
    Each piece is translated: the cube root becomes t1/3t^{1/3}⁠, the spare x3x^3 becomes t+1t+1⁠, and x2 dxx^2\,dx becomes dt3\frac{dt}{3}⁠. As xx runs over all reals, so does tt⁠.
    Substitution, Section 7.3.1; Property (IV), Section 7.2.1. Lesson: integration by substitution

  3. Expand into powers of tt

    (t+1) t1/3=t4/3+t1/3.(t+1)\,t^{1/3} = t^{4/3} + t^{1/3}.
    The power formula integrates a single power of tt⁠, not a product, so we multiply out first. Multiplying powers of the same base adds the exponents: 1+13=431 + \frac13 = \frac43⁠.
    t⋅t1/3=t1+1/3=t4/3t\cdot t^{1/3} = t^{1 + 1/3} = t^{4/3}⁠. Basics: powers of x

  4. Integrate term by term

    13∫(t4/3+t1/3)dt=13[t7/37/3+t4/34/3]+C=13[37 t7/3+34 t4/3]+C=17 t7/3+14 t4/3+C.\begin{aligned} \frac13\int\left(t^{4/3} + t^{1/3}\right)dt &= \frac13\left[\frac{t^{7/3}}{7/3} + \frac{t^{4/3}}{4/3}\right] + C \\ &= \frac13\left[\frac37\,t^{7/3} + \frac34\,t^{4/3}\right] + C \\ &= \frac17\,t^{7/3} + \frac14\,t^{4/3} + C. \end{aligned}
    Power formula with n=43n = \frac43 (new power 73\frac73⁠) and n=13n = \frac13 (new power 43\frac43⁠). The 13\frac13 outside cancels the 33 in each numerator. Here t7/3t^{7/3} means (t1/3)7(t^{1/3})^7 and t4/3t^{4/3} means (t1/3)4(t^{1/3})^4⁠, which are defined for negative tt too.
    Standard formula (i), Section 7.2; Property (V), Section 7.2.1.

  5. Return to xx

    17 t7/3+14 t4/3+C=17(x3−1)7/3+14(x3−1)4/3+C.\begin{aligned} &\frac17\,t^{7/3} + \frac14\,t^{4/3} + C \\ &\quad = \frac17\left(x^3-1\right)^{7/3} + \frac14\left(x^3-1\right)^{4/3} + C. \end{aligned}
    Replace tt by x3−1x^3 - 1 everywhere. This is exactly the form in the answer key.

  6. Check: differentiate the answer

    Let F(x)F(x) be the answer. Chain rule with inner u=x3−1u = x^3-1 (u′=3x2u' = 3x^2⁠):

    F′(x)=17⋅73 u4/3⋅3x2+14⋅43 u1/3⋅3x2=x2 u4/3+x2 u1/3=x2 u1/3 (u+1).\begin{aligned} F'(x) &= \frac17\cdot\frac73\,u^{4/3}\cdot 3x^2 + \frac14\cdot\frac43\,u^{1/3}\cdot 3x^2 \\ &= x^2\,u^{4/3} + x^2\,u^{1/3} \\ &= x^2\,u^{1/3}\,(u + 1). \end{aligned}

    Since u+1=x3u + 1 = x^3⁠, this is x2⋅x3 (x3−1)1/3=(x3−1)1/3x5x^2\cdot x^3\,(x^3-1)^{1/3} = (x^3-1)^{1/3}x^5⁠, the integrand.

    The constants cancel exactly: 17⋅73⋅3=1\frac17\cdot\frac73\cdot 3 = 1 and 14⋅43⋅3=1\frac14\cdot\frac43\cdot 3 = 1⁠. The factoring in the last line is Step 3 run backwards. (At the single point x=1x = 1⁠, where u=0u = 0⁠, a direct limit shows F′(1)=0F'(1) = 0⁠, which also matches the integrand.)
    Chain rule, Chapter 5, Theorem 4. Basics: integration undoes differentiation
Result

For every real xx (real cube roots),

∫(x3−1)1/3x5 dx=17(x3−1)7/3+14(x3−1)4/3+C.\begin{aligned} &\int(x^3-1)^{1/3}x^5\,dx \\ &\quad = \frac17(x^3-1)^{7/3} + \frac14(x^3-1)^{4/3} + C. \end{aligned}

(Textbook answer: 17(x3−1)7/3+14(x3−1)4/3+C\frac17(x^3-1)^{7/3} + \frac14(x^3-1)^{4/3} + C⁠.)

Check it with numbers

At x=2x = 2 the integrand is 71/3⋅32≈61.21387^{1/3}\cdot 32 \approx 61.2138⁠. With h=0.0001h = 0.0001⁠:

F(2.0001)≈16.74427017,F(1.9999)≈16.73202741,F(2.0001)−F(1.9999)0.0002≈61.2138.\begin{aligned} F(2.0001) &\approx 16.74427017, \\ F(1.9999) &\approx 16.73202741, \\ \frac{F(2.0001)-F(1.9999)}{0.0002} &\approx 61.2138. \end{aligned}

The slope of FF equals the integrand to four decimal places (the small remaining gap is the error of the difference quotient, which shrinks as hh does).

Common mistakes
  1. Using all of x5 dxx^5\,dx for dtdt⁠. Only x2 dxx^2\,dx is part of dtdt⁠; the other x3x^3 must become t+1t+1⁠.
  2. Writing x3=t−1x^3 = t - 1⁠. From t=x3−1t = x^3 - 1⁠, x3=t+1x^3 = t + 1⁠.
  3. Arithmetic with the fractional powers. Dividing by 73\frac73 multiplies by 37\frac37⁠; then 13⋅37=17\frac13\cdot\frac37 = \frac17⁠.
What to take away
  1. When the power of xx outside is larger than the derivative needs, split it: one part for dtdt⁠, the rest rewritten through tt⁠.
  2. The same idea handles ∫x5x3+a dx\int x^5\sqrt{x^3+a}\,dx and similar integrals.

Every fact used, and where it comes from

FactUsed inSource
ddx(x3−1)=3x2\frac{d}{dx}(x^3-1) = 3x^2Steps 1, 2Chapter 5, Table 5.3
Laws of exponentsSteps 1, 3, 6Algebra (Basics)
SubstitutionStep 2Section 7.3.1 (Lesson: substitution)
∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 4Standard formula (i), Section 7.2
Integral of a sum; constant factorsSteps 2, 4Properties (IV), (V), Section 7.2.1
Chain ruleStep 6Chapter 5, Theorem 4
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