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Chapter 7 · Integrals

Exercise 7.2 · Q13: ∫x2(2+3x3)3 dx\int \frac{x^2}{(2+3x^3)^3}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function:

x2(2+3x3)3\frac{x^2}{\left(2+3x^3\right)^3}
Exam answer

Let I=∫x2(2+3x3)3 dxI = \displaystyle\int\frac{x^2}{\left(2 + 3x^3\right)^3}\,dx⁠.

Put 2+3x3=t2 + 3x^3 = t⁠. Then 9x2 dx=dt9x^2\,dx = dt⁠.

I=19∫dtt3=19⋅t−2−2+C=−118 t2+C\begin{aligned} I &= \frac19\int\frac{dt}{t^3} = \frac19\cdot\frac{t^{-2}}{-2} + C \\ &= -\frac{1}{18\,t^2} + C \end{aligned}

∴ I=−118(2+3x3)2+C\therefore\ I = -\dfrac{1}{18\left(2 + 3x^3\right)^2} + C⁠.

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

The one idea this problem is built on

The bracket 2+3x32+3x^3 has derivative 9x29x^2⁠, and the numerator is x2x^2⁠. So t=2+3x3t = 2+3x^3 gives x2 dx=dt9x^2\,dx = \frac{dt}{9}⁠, and the integral is 19∫t−3 dt\frac19\int t^{-3}\,dt⁠: a negative power, handled by the same power formula.

Given, to find, and the plan
Given
f(x)=x2(2+3x3)3f(x) = \dfrac{x^2}{(2+3x^3)^3}⁠. It is defined wherever 2+3x3≠02 + 3x^3 \ne 0⁠, that is x3≠−23x^3 \ne -\frac23⁠, or x≠−(23)1/3≈−0.8736x \ne -\left(\frac23\right)^{1/3} \approx -0.8736⁠.
To find
∫x2(2+3x3)3 dx\displaystyle\int\frac{x^2}{(2+3x^3)^3}\,dx⁠.
Plan
(1) Differentiate the bracket and compare with the numerator. (2) Substitute t=2+3x3t = 2+3x^3⁠. (3) Power formula with n=−3n = -3⁠. (4) Return to xx⁠. (5) Differentiate to check.

Basics you need

1Negative powers

1t3=t−3\frac{1}{t^3} = t^{-3} and t−2=1t2t^{-2} = \frac{1}{t^2}⁠, for t≠0t \ne 0⁠. Basics: powers of x

2Substitution

Put t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dx (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 for every n≠−1n \ne -1⁠, including negative nn (standard formula (i), Section 7.2).

Solution

  1. Compare the numerator with the bracket's derivative

    ddx(2+3x3)=9x2.\frac{d}{dx}\left(2 + 3x^3\right) = 9x^2.

    The numerator x2x^2 is 19\frac19 of this.

    The derivative of the inner expression is present up to a constant factor, so t=2+3x3t = 2 + 3x^3 is the substitution to use.

  2. Substitute t=2+3x3t = 2 + 3x^3

    t=2+3x3,dt=9x2 dx,x2 dx=dt9.\begin{aligned} t &= 2 + 3x^3, \\ dt &= 9x^2\,dx, \\ x^2\,dx &= \frac{dt}{9}. \end{aligned}

    Therefore

    ∫x2 dx(2+3x3)3=19∫dtt3=19∫t−3 dt.\int\frac{x^2\,dx}{\left(2+3x^3\right)^3} = \frac19\int\frac{dt}{t^3} = \frac19\int t^{-3}\,dt.
    The numerator together with dxdx becomes dt9\frac{dt}{9}⁠, and the bracket becomes tt⁠. On the domain, t≠0t \ne 0⁠, so t−3t^{-3} is defined.
    Substitution, Section 7.3.1; Property (IV), Section 7.2.1. Lesson: integration by substitution

  3. Integrate the negative power

    19∫t−3 dt=19⋅t−2−2+C=−118 t2+C.\begin{aligned} \frac19\int t^{-3}\,dt &= \frac19\cdot\frac{t^{-2}}{-2} + C \\ &= -\frac{1}{18\,t^2} + C. \end{aligned}
    The power formula with n=−3n = -3 (allowed: it is not −1-1⁠): the new power is −3+1=−2-3 + 1 = -2⁠, and we divide by −2-2⁠. Then 19⋅1−2=−118\frac19\cdot\frac{1}{-2} = -\frac{1}{18}⁠, and t−2=1t2t^{-2} = \frac{1}{t^2}⁠.
    Standard formula (i), Section 7.2. Basics: the power rule

  4. Return to xx

    −118 t2+C=−118(2+3x3)2+C.-\frac{1}{18\,t^2} + C = -\frac{1}{18\left(2+3x^3\right)^2} + C.
    Replace tt by 2+3x32+3x^3 so that the answer is in xx⁠. It is defined exactly where the integrand is, namely where 2+3x3≠02+3x^3 \ne 0⁠.

  5. Check: differentiate the answer

    Write the answer as F(x)=−118 u−2F(x) = -\frac{1}{18}\,u^{-2} with u=2+3x3u = 2+3x^3 (u′=9x2u' = 9x^2⁠). By the chain rule,

    F′(x)=−118⋅(−2) u−3⋅9x2=18x218 u3=x2(2+3x3)3.\begin{aligned} F'(x) &= -\frac{1}{18}\cdot(-2)\,u^{-3}\cdot 9x^2 \\ &= \frac{18x^2}{18\,u^3} \\ &= \frac{x^2}{\left(2+3x^3\right)^3}. \end{aligned}

    We are back at the integrand, wherever 2+3x3≠02 + 3x^3 \ne 0⁠.

    The power −2-2 brought down by the power rule turns −118-\frac{1}{18} into 19\frac19⁠, and the inner derivative 9x29x^2 cancels the 19\frac19⁠. Getting the integrand back confirms the answer (Property (I), Section 7.2.1).
    Chain rule, Chapter 5, Theorem 4; dduu−2=−2u−3\frac{d}{du}u^{-2} = -2u^{-3}⁠. Basics: integration undoes differentiation
Result

For x≠−(23)1/3x \ne -\left(\frac23\right)^{1/3}⁠,

∫x2(2+3x3)3 dx=−118(2+3x3)2+C.\int\frac{x^2}{\left(2+3x^3\right)^3}\,dx = -\frac{1}{18\left(2+3x^3\right)^2} + C.

(Textbook answer: −118(2+3x3)2+C-\frac{1}{18(2+3x^3)^2} + C⁠.) The constant may be chosen separately on each side of x=−(23)1/3x = -\left(\frac23\right)^{1/3}⁠.

Check it with numbers

At x=1x = 1 the integrand is 153=0.008\frac{1}{5^3} = 0.008⁠. With h=0.0001h = 0.0001⁠:

F(1.0001)≈−0.00222142,F(0.9999)≈−0.00222302,F(1.0001)−F(0.9999)0.0002≈0.008000.\begin{aligned} F(1.0001) &\approx -0.00222142, \\ F(0.9999) &\approx -0.00222302, \\ \frac{F(1.0001)-F(0.9999)}{0.0002} &\approx 0.008000. \end{aligned}

The slope of FF equals the integrand.

Common mistakes
  1. Using log⁡\log for a power in the denominator. ∫dtt3\int\frac{dt}{t^3} is not log⁡∣t3∣\log|t^3|⁠; the log formula is only for 1t\frac1t (power −1-1⁠).
  2. Getting the new power wrong. −3+1=−2-3 + 1 = -2⁠, not −4-4⁠.
  3. Losing the minus sign. Dividing by −2-2 makes the answer negative.
  4. Wrong constant. x2 dx=dt9x^2\,dx = \frac{dt}{9}⁠, since ddx(3x3)=9x2\frac{d}{dx}(3x^3) = 9x^2⁠.
What to take away
  1. The power formula works for negative powers too, except −1-1⁠, which gives a log.
  2. In general, for n≠−1n \ne -1⁠, ∫g′(x) g(x)n dx=g(x)n+1n+1+C.\int g'(x)\,g(x)^{n}\,dx = \frac{g(x)^{n+1}}{n+1} + C.

Every fact used, and where it comes from

FactUsed inSource
ddx(2+3x3)=9x2\frac{d}{dx}(2+3x^3) = 9x^2Steps 1, 2Chapter 5, Table 5.3
SubstitutionStep 2Section 7.3.1 (Lesson: substitution)
1t3=t−3\frac{1}{t^3} = t^{-3}Steps 2, 3Laws of exponents
∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 3Standard formula (i), Section 7.2
Chain ruleStep 5Chapter 5, Theorem 4
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