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Chapter 7 · Integrals

Exercise 7.7 · Q9: ∫1+x29 dx\int\sqrt{1 + \frac{x^2}{9}}\,dx

Exercise 7.7Integrals of some more types · Section 7.6.2
The question

Integrate the function

1+x29.\sqrt{1 + \frac{x^2}{9}}.
Exam answer

Let I=∫1+x29 dxI = \displaystyle\int\sqrt{1 + \frac{x^2}{9}}\,dx⁠.

1+x29=13x2+32\sqrt{1 + \frac{x^2}{9}} = \frac13\sqrt{x^2 + 3^2}

Using ∫x2+a2 dx=x2x2+a2\displaystyle\int\sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} +a22log⁡∣x+x2+a2∣+C,+ \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C,

I=13[x2x2+9+92log⁡∣x+x2+9∣]+C\begin{aligned} I &= \frac13\bigg[\frac{x}{2}\sqrt{x^2 + 9} \\ &\quad + \frac92\log\left|x + \sqrt{x^2 + 9}\right|\bigg] + C \end{aligned}

∴ I=x6x2+9\therefore\ I = \dfrac{x}{6}\sqrt{x^2 + 9} +32log⁡∣x+x2+9∣+C+ \dfrac32\log\left|x + \sqrt{x^2 + 9}\right| + C⁠.

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

The one idea this problem is built on

Put the inside over a common denominator and take the square root of the 9 out:

1+x29=x2+99=13x2+32.\sqrt{1 + \frac{x^2}{9}} = \sqrt{\frac{x^2 + 9}{9}} = \frac13\sqrt{x^2 + 3^2}.

Then formula 7.6.2 (ii) with a=3a = 3 does the rest, and the 13\frac13 multiplies the whole answer.

Given, to find, and the plan
Given
The integrand 1+x29\sqrt{1 + \frac{x^2}{9}}⁠.
To find
∫1+x29 dx\int\sqrt{1 + \frac{x^2}{9}}\,dx⁠, and where the answer is valid.
Plan
(1) Note the domain. (2) Rewrite as 13x2+9\frac13\sqrt{x^2 + 9}⁠. (3) Apply 7.6.2 (ii) with a=3a = 3⁠. (4) Multiply by 13\frac13⁠. (5) Differentiate back.

Basics you need

1Roots of quotients

For p≥0p \ge 0 and q>0q > 0⁠, pq=pq\sqrt{\frac{p}{q}} = \frac{\sqrt p}{\sqrt q}⁠. So x2+99=x2+93\sqrt{\frac{x^2 + 9}{9}} = \frac{\sqrt{x^2 + 9}}{3}⁠. Basics: powers and roots

2Formula 7.6.2 (ii)

For a>0a > 0⁠, ∫x2+a2 dx\int\sqrt{x^2 + a^2}\,dx equals x2x2+a2+a22log⁡∣x+x2+a2∣\frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right|⁠, plus CC⁠. Derived by parts in Exercise 7.7 Q3. Lesson: integration by parts

3Constant multiples

∫kf(x) dx=k∫f(x) dx\int k f(x)\,dx = k\int f(x)\,dx (Property (IV), Section 7.2.1). Basics: integration undoes differentiation

Solution

  1. The domain: every real xx

    1+x29≥1>01 + \frac{x^2}{9} \ge 1 > 0 for every real xx⁠, so the integrand is defined everywhere.

    x2≥0x^2 \ge 0⁠, so adding x29\frac{x^2}{9} to 1 never makes it smaller than 1.

  2. Take the 9 out of the root

    1+x29=9+x29=x2+99=13x2+9.\begin{aligned} \sqrt{1 + \frac{x^2}{9}} &= \sqrt{\frac{9 + x^2}{9}} = \frac{\sqrt{x^2 + 9}}{\sqrt9} \\ &= \frac13\sqrt{x^2 + 9}. \end{aligned}
    1=991 = \frac99⁠, so 1+x29=9+x291 + \frac{x^2}{9} = \frac{9 + x^2}{9}⁠. The root of a quotient of positive numbers is the quotient of the roots, and 9=3\sqrt9 = 3⁠. No completing the square is needed: there is no xx term.
    Laws of roots. Basics: powers and roots

  3. Apply 7.6.2 (ii) with a=3a = 3

    With a2=9a^2 = 9⁠, a22=92\frac{a^2}{2} = \frac92⁠:

    ∫x2+9 dx=x2x2+9+92log⁡∣x+x2+9∣+C1.\begin{aligned} \int\sqrt{x^2 + 9}\,dx &= \frac{x}{2}\sqrt{x^2 + 9} \\ &\quad + \frac92\log\left|x + \sqrt{x^2 + 9}\right| + C_1. \end{aligned}
    The root is "x2x^2 plus a positive number", so it is formula (ii), and the log term is added.
    Formula 7.6.2 (ii). Lesson: integration by parts

  4. Multiply by 13\frac13

    ∫1+x29 dx=13∫x2+9 dx=x6x2+9+32log⁡∣x+x2+9∣+C.\begin{aligned} \int\sqrt{1 + \tfrac{x^2}{9}}\,dx &= \frac13\int\sqrt{x^2 + 9}\,dx \\ &= \frac{x}{6}\sqrt{x^2 + 9} \\ &\quad + \frac32\log\left|x + \sqrt{x^2 + 9}\right| + C. \end{aligned}
    13⋅x2=x6\frac13\cdot\frac{x}{2} = \frac{x}{6} and 13⋅92=32\frac13\cdot\frac92 = \frac32⁠; C=C13C = \frac{C_1}{3} is still arbitrary. Here x+x2+9>0x + \sqrt{x^2 + 9} > 0 for every xx (the root exceeds ∣x∣|x|⁠), so the modulus could be dropped; the textbook keeps it.
    Property (IV), Section 7.2.1. Basics: integration undoes differentiation

  5. Check: differentiate the answer

    Write r=x2+9r = \sqrt{x^2 + 9}⁠, so drdx=xr\frac{dr}{dx} = \frac{x}{r}⁠. Then

    ddx[x6 r]=r6+x26r,ddx[32log⁡∣x+r∣]=32⋅1+xrx+r=32r=96r.\begin{aligned} \frac{d}{dx}\left[\frac{x}{6}\,r\right] &= \frac{r}{6} + \frac{x^2}{6r}, \\ \frac{d}{dx}\left[\frac32\log|x + r|\right] &= \frac32\cdot\frac{1 + \frac{x}{r}}{x + r} \\ &= \frac{3}{2r} = \frac{9}{6r}. \end{aligned}

    Adding, over 6r6r⁠:

    r2+x2+96r=2r26r=r3=x2+93=1+x29. ✓\begin{aligned} \frac{r^2 + x^2 + 9}{6r} &= \frac{2r^2}{6r} = \frac{r}{3} \\ &= \frac{\sqrt{x^2 + 9}}{3} = \sqrt{1 + \frac{x^2}{9}}.\ \checkmark \end{aligned}
    r2=x2+9r^2 = x^2 + 9⁠, so the numerator is 2r22r^2⁠. The last equality is Step 2 read backwards. The check holds for every real xx⁠.
    Product rule; chain rule; standard formula (xii), Section 7.2. Basics: logarithms and 1/x
Result ∫1+x29 dx=x6x2+9+32log⁡∣x+x2+9∣+C,\begin{aligned} \int\sqrt{1 + \tfrac{x^2}{9}}\,dx &= \frac{x}{6}\sqrt{x^2 + 9} \\ &\quad + \frac32\log\left|x + \sqrt{x^2 + 9}\right| + C, \end{aligned}

valid for every real xx⁠. (Textbook answer: x6x2+9\frac{x}{6}\sqrt{x^2 + 9} +32log⁡∣x+x2+9∣+C+ \frac32\log\left|x + \sqrt{x^2 + 9}\right| + C⁠, the same.)

Check it with numbers

By Section 7.8, ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) for our answer FF⁠, on any interval where the integrand is continuous. The middle column is a computer's sum of very many thin strips, found without our answer.

IntervalSum of thin stripsF(b)−F(a)F(b) - F(a)
[0,3][0, 3]3.44338073.44338073.44338073.4433807
[−2,4][-2, 4]7.12081997.12081997.12081997.1208199
Common mistakes
  1. Treating x29+1\sqrt{\frac{x^2}{9} + 1} as if it were x2+a2\sqrt{x^2 + a^2}⁠. The x2x^2 carries the coefficient 19\frac19⁠. Take it out first (Step 2), or substitute t=x3t = \frac{x}{3} with dx=3 dtdx = 3\,dt⁠; applying the formula directly gives wrong factors.
  2. Forgetting the 13\frac13⁠. Without it the answer is three times too big: x2x2+9+92log⁡…\frac{x}{2}\sqrt{x^2 + 9} + \frac92\log\dots⁠.
  3. 1+x29=1+x3\sqrt{1 + \frac{x^2}{9}} = 1 + \frac{x}{3}⁠. A root of a sum is not the sum of the roots.
What to take away
  1. Clear denominators inside a root first: pq=pq\sqrt{\frac{p}{q}} = \frac{\sqrt p}{\sqrt q}⁠.
  2. The substitution t=x3t = \frac{x}{3} turns the integral into 3∫1+t2 dt3\int\sqrt{1 + t^2}\,dt⁠. Back in xx its log term is 32log⁡∣x+x2+93∣\frac32\log\left|\frac{x + \sqrt{x^2 + 9}}{3}\right|⁠, which differs from ours only by the constant 32log⁡3\frac32\log 3⁠.

Every fact used, and where it comes from

FactUsed inSource
p/q=p/q\sqrt{p/q} = \sqrt p/\sqrt q for p≥0p \ge 0⁠, q>0q > 0Step 2Laws of roots (Basics: powers and roots)
∫x2+a2 dx\int\sqrt{x^2 + a^2}\,dx formulaStep 3Formula 7.6.2 (ii), Section 7.6.2
∫kf(x) dx=k∫f(x) dx\int k f(x)\,dx = k\int f(x)\,dxStep 4Property (IV), Section 7.2.1
ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u}Step 5Standard formula (xii), Section 7.2
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Numerical checkSection 7.8 (Lesson: definite integrals)
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