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Chapter 7 · Integrals

Exercise 7.2 · Q9: ∫(4x+2)x2+x+1 dx\int (4x+2)\sqrt{x^2+x+1}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function:

(4x+2)x2+x+1(4x+2)\sqrt{x^2+x+1}
Exam answer

Let I=∫(4x+2)x2+x+1 dxI = \displaystyle\int(4x + 2)\sqrt{x^2 + x + 1}\,dx⁠.

I=2∫(2x+1)x2+x+1 dxI = 2\int(2x + 1)\sqrt{x^2 + x + 1}\,dx

Put x2+x+1=tx^2 + x + 1 = t⁠. Then (2x+1) dx=dt(2x + 1)\,dx = dt⁠.

∴ I=2∫t1/2 dt=2⋅t3/23/2+C\begin{aligned} \therefore\ I &= 2\int t^{1/2}\,dt \\ &= 2\cdot\frac{t^{3/2}}{3/2} + C \end{aligned}

∴ I=43(x2+x+1)3/2+C\therefore\ I = \dfrac43\left(x^2 + x + 1\right)^{3/2} + C⁠.

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

The one idea this problem is built on

The derivative of x2+x+1x^2+x+1 is 2x+12x+1⁠, and the factor in front is 4x+2=2(2x+1)4x+2 = 2(2x+1)⁠. So with t=x2+x+1t = x^2+x+1⁠, the factor (4x+2) dx(4x+2)\,dx is exactly 2 dt2\,dt⁠, and the integral becomes 2∫t dt2\int\sqrt t\,dt⁠.

Given, to find, and the plan
Given
f(x)=(4x+2)x2+x+1f(x) = (4x+2)\sqrt{x^2+x+1}⁠. Completing the square, x2+x+1=(x+12)2+34≥34>0x^2+x+1 = \left(x+\frac12\right)^2 + \frac34 \ge \frac34 > 0⁠, so ff is defined and continuous for every real xx⁠.
To find
∫(4x+2)x2+x+1 dx\displaystyle\int (4x+2)\sqrt{x^2+x+1}\,dx⁠.
Plan
(1) Differentiate the inside and compare with 4x+24x+2⁠. (2) Substitute t=x2+x+1t = x^2+x+1⁠. (3) Power formula. (4) Return to xx⁠. (5) Differentiate to check.

Basics you need

1Completing the square

x2+x+1=(x+12)2+34x^2+x+1 = \left(x+\frac12\right)^2 + \frac34⁠, which shows it is always positive. Basics: completing the square

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, n≠−1n \ne -1 (standard formula (i), Section 7.2). Basics: the power rule

Solution

  1. Compare the front factor with the inside's derivative

    ddx(x2+x+1)=2x+1,4x+2=2(2x+1).\begin{aligned} \frac{d}{dx}\left(x^2+x+1\right) &= 2x + 1, \\ 4x + 2 &= 2(2x+1). \end{aligned}
    The front factor is twice the derivative of the expression under the root. That is the signal for the substitution t=x2+x+1t = x^2+x+1⁠.

  2. Substitute t=x2+x+1t = x^2+x+1

    t=x2+x+1,dt=(2x+1) dx,(4x+2) dx=2 dt.\begin{aligned} t &= x^2 + x + 1, \\ dt &= (2x+1)\,dx, \\ (4x+2)\,dx &= 2\,dt. \end{aligned}

    Therefore

    ∫x2+x+1 (4x+2) dx=∫t⋅2 dt=2∫t1/2 dt.\int\sqrt{x^2+x+1}\,(4x+2)\,dx = \int\sqrt t\cdot 2\,dt = 2\int t^{1/2}\,dt.
    Multiplying dt=(2x+1) dxdt = (2x+1)\,dx by 2 gives exactly the piece (4x+2) dx(4x+2)\,dx in the integrand. Since t≥34>0t \ge \frac34 > 0⁠, t=t1/2\sqrt t = t^{1/2} is defined.
    Substitution, Section 7.3.1. Lesson: integration by substitution

  3. Integrate the power

    2∫t1/2 dt=2⋅t3/23/2+C=43 t3/2+C.2\int t^{1/2}\,dt = 2\cdot\frac{t^{3/2}}{3/2} + C = \frac43\,t^{3/2} + C.
    The power formula with n=12n = \frac12⁠: the new power is 32\frac32⁠, and dividing by 32\frac32 means multiplying by 23\frac23⁠. Then 2⋅23=432\cdot\frac23 = \frac43⁠.
    Standard formula (i), Section 7.2, with n=12n = \frac12⁠.

  4. Return to xx

    43 t3/2+C=43(x2+x+1)3/2+C.\frac43\,t^{3/2} + C = \frac43\left(x^2+x+1\right)^{3/2} + C.
    Replace tt by x2+x+1x^2+x+1 so that the answer is in xx⁠. Because x2+x+1≥34>0x^2+x+1 \ge \frac34 > 0⁠, the power 32\frac32 is defined for every real xx⁠.

  5. Check: differentiate the answer

    Let F(x)=43(x2+x+1)3/2F(x) = \frac43(x^2+x+1)^{3/2}⁠. Chain rule with inner u=x2+x+1u = x^2+x+1 (u′=2x+1u' = 2x+1⁠):

    F′(x)=43⋅32(x2+x+1)1/2(2x+1)=2(2x+1)x2+x+1=(4x+2)x2+x+1.\begin{aligned} F'(x) &= \frac43\cdot\frac32\left(x^2+x+1\right)^{1/2}(2x+1) \\ &= 2(2x+1)\sqrt{x^2+x+1} \\ &= (4x+2)\sqrt{x^2+x+1}. \end{aligned}

    This is the integrand, for every real xx⁠.

    The power rule's 32\frac32 cancels the 23\frac23 inside 43\frac43⁠, leaving 22⁠; times the inner derivative 2x+12x+1 this rebuilds the factor 4x+24x+2⁠. Getting the integrand back confirms the answer (Property (I), Section 7.2.1).
    Chain rule, Chapter 5, Theorem 4. Basics: integration undoes differentiation
Result

For every real xx⁠,

∫(4x+2)x2+x+1 dx=43(x2+x+1)3/2+C.\int (4x+2)\sqrt{x^2+x+1}\,dx = \frac43\left(x^2+x+1\right)^{3/2} + C.

(Textbook answer: 43(x2+x+1)3/2+C\frac43(x^2+x+1)^{3/2} + C⁠.)

Check it with numbers

At x=1x = 1 the integrand is 63≈10.3923056\sqrt3 \approx 10.392305⁠. With h=0.0001h = 0.0001⁠:

F(1.0001)≈6.929242521,F(0.9999)≈6.927164060,F(1.0001)−F(0.9999)0.0002≈10.392305.\begin{aligned} F(1.0001) &\approx 6.929242521, \\ F(0.9999) &\approx 6.927164060, \\ \frac{F(1.0001)-F(0.9999)}{0.0002} &\approx 10.392305. \end{aligned}

The slope of FF equals the integrand.

Common mistakes
  1. Missing the factor 2. (4x+2) dx=2 dt(4x+2)\,dx = 2\,dt⁠, not dtdt⁠; forgetting it halves the answer.
  2. Expanding the product. Multiplying 4x+24x+2 into the root does not help; the substitution does all the work.
  3. Worrying about the root's domain. x2+x+1x^2+x+1 is never negative (complete the square), so no restriction is needed here.
What to take away
  1. Factor the front expression to see whether it is a multiple of the inside's derivative.
  2. ∫g′(x)g(x) dx=23 g(x)3/2+C\int g'(x)\sqrt{g(x)}\,dx = \frac23\,g(x)^{3/2} + C wherever g(x)≥0g(x) \ge 0⁠.

Every fact used, and where it comes from

FactUsed inSource
x2+x+1=(x+12)2+34x^2+x+1 = (x+\frac12)^2 + \frac34Given, Step 2Completing the square (Basics)
ddx(x2+x+1)=2x+1\frac{d}{dx}(x^2+x+1) = 2x+1Step 1Chapter 5, Table 5.3
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 3Standard formula (i), Section 7.2
Chain ruleStep 5Chapter 5, Theorem 4
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