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Chapter 7 · Integrals

Exercise 7.2 · Q8: ∫x1+2x2 dx\int x\sqrt{1+2x^2}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function:

x1+2x2x\sqrt{1+2x^2}
Exam answer

Let I=∫x1+2x2 dxI = \displaystyle\int x\sqrt{1 + 2x^2}\,dx⁠.

Put 1+2x2=t1 + 2x^2 = t⁠. Then 4x dx=dt4x\,dx = dt⁠.

∴ I=14∫t1/2 dt=14⋅t3/23/2+C\begin{aligned} \therefore\ I &= \frac14\int t^{1/2}\,dt \\ &= \frac14\cdot\frac{t^{3/2}}{3/2} + C \end{aligned}

∴ I=16(1+2x2)3/2+C\therefore\ I = \dfrac16\left(1 + 2x^2\right)^{3/2} + C⁠.

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

The one idea this problem is built on

The inside of the root, 1+2x21+2x^2⁠, has derivative 4x4x⁠, and the factor outside is xx⁠: the same thing up to the constant 44⁠. So t=1+2x2t = 1+2x^2 gives x dx=dt4x\,dx = \frac{dt}{4}⁠, and the integral is 14∫t dt\frac14\int\sqrt t\,dt⁠.

Given, to find, and the plan
Given
f(x)=x1+2x2f(x) = x\sqrt{1+2x^2}⁠. Since 1+2x2≥1>01+2x^2 \ge 1 > 0⁠, ff is defined and continuous for every real xx⁠.
To find
∫x1+2x2 dx\displaystyle\int x\sqrt{1+2x^2}\,dx⁠.
Plan
(1) Compare the outside factor with the derivative of the inside. (2) Substitute t=1+2x2t = 1+2x^2⁠. (3) Power formula. (4) Return to xx⁠. (5) Differentiate to check.

Basics you need

1Substitution with a constant to spare

If the integrand contains k⋅g′(x)k\cdot g'(x) instead of exactly g′(x)g'(x)⁠, the substitution t=g(x)t = g(x) still works: adjust by the constant. Basics: substitution

2The 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. Differentiate the inside of the root

    ddx(1+2x2)=4x.\frac{d}{dx}\left(1+2x^2\right) = 4x.

    The integrand contains xx⁠, which is 14\frac14 of this.

    Substitution needs the derivative of the inner function to be present, possibly times a constant. Here the constant is 14\frac14⁠, so t=1+2x2t = 1+2x^2 will work.

  2. Substitute t=1+2x2t = 1+2x^2

    t=1+2x2,dt=4x dx,x dx=dt4.\begin{aligned} t &= 1 + 2x^2, \\ dt &= 4x\,dx, \\ x\,dx &= \frac{dt}{4}. \end{aligned}

    Therefore

    ∫1+2x2 (x dx)=∫t⋅dt4=14∫t1/2 dt.\int\sqrt{1+2x^2}\,(x\,dx) = \int\sqrt t\cdot\frac{dt}{4} = \frac14\int t^{1/2}\,dt.
    We solve dt=4x dxdt = 4x\,dx for the exact piece in the integrand, x dxx\,dx⁠. Then nothing in xx remains. Since t≥1t \ge 1⁠, t=t1/2\sqrt t = t^{1/2} is defined.
    Substitution, Section 7.3.1; Property (IV), Section 7.2.1. Lesson: integration by substitution

  3. Integrate the power

    14∫t1/2 dt=14⋅t3/23/2+C=16 t3/2+C.\frac14\int t^{1/2}\,dt = \frac14\cdot\frac{t^{3/2}}{3/2} + C = \frac16\,t^{3/2} + C.
    Power formula with n=12n = \frac12⁠; then 14⋅23=16\frac14\cdot\frac23 = \frac16⁠.
    Standard formula (i), Section 7.2.

  4. Return to xx

    16 t3/2+C=16(1+2x2)3/2+C.\frac16\,t^{3/2} + C = \frac16\left(1+2x^2\right)^{3/2} + C.
    The answer must be in the original variable, so tt is replaced by 1+2x21+2x^2⁠. Since 1+2x2≥11+2x^2 \ge 1⁠, the power 32\frac32 is defined for every real xx and no restriction is needed.

  5. Check: differentiate the answer

    Let F(x)=16(1+2x2)3/2F(x) = \frac16(1+2x^2)^{3/2}⁠. Chain rule with inner u=1+2x2u = 1+2x^2 (u′=4xu' = 4x⁠):

    F′(x)=16⋅32(1+2x2)1/2⋅4x=x1+2x2.\begin{aligned} F'(x) &= \frac16\cdot\frac32\left(1+2x^2\right)^{1/2}\cdot 4x \\ &= x\sqrt{1+2x^2}. \end{aligned}

    Indeed 16⋅32⋅4=1\frac16\cdot\frac32\cdot 4 = 1⁠. We are back at the integrand, for every real xx⁠.

    An answer is right exactly when its derivative is the integrand (Property (I), Section 7.2.1). The 14\frac14 from x dx=dt4x\,dx = \frac{dt}{4} and the 23\frac23 from the power formula are undone by the inner derivative 4x4x and the power rule's 32\frac32⁠.
    Chain rule, Chapter 5, Theorem 4. Basics: integration undoes differentiation
Result

For every real xx⁠,

∫x1+2x2 dx=16(1+2x2)3/2+C.\int x\sqrt{1+2x^2}\,dx = \frac16\left(1+2x^2\right)^{3/2} + C.

(Textbook answer: 16(1+2x2)3/2+C\frac16(1+2x^2)^{3/2} + C⁠.)

Check it with numbers

At x=2x = 2 the integrand is 29=62\sqrt{9} = 6⁠. With h=0.0001h = 0.0001⁠:

F(2.0001)≈4.50060003,F(1.9999)≈4.49940003,F(2.0001)−F(1.9999)0.0002≈6.000000.\begin{aligned} F(2.0001) &\approx 4.50060003, \\ F(1.9999) &\approx 4.49940003, \\ \frac{F(2.0001)-F(1.9999)}{0.0002} &\approx 6.000000. \end{aligned}

The slope of FF equals the integrand.

Common mistakes
  1. Writing x dx=4 dtx\,dx = 4\,dt⁠. From dt=4x dxdt = 4x\,dx⁠, divide by 4: x dx=dt4x\,dx = \frac{dt}{4}⁠.
  2. Forgetting to divide by the new power 32\frac32⁠. The answer would be off by the factor 32\frac32⁠.
  3. Substituting t=x2t = x^2 only. It works too but leaves 1+2t\sqrt{1+2t}⁠, needing a second substitution.
What to take away
  1. If the outside factor is a constant multiple of the inside's derivative, substitute and fix the constant.
  2. Always solve dt=g′(x) dxdt = g'(x)\,dx for the exact combination that appears in the integrand.

Every fact used, and where it comes from

FactUsed inSource
ddx(1+2x2)=4x\frac{d}{dx}(1+2x^2) = 4xSteps 1, 2Chapter 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
Constant factors move outsideStep 2Property (IV), Section 7.2.1
Chain ruleStep 5Chapter 5, Theorem 4
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