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Chapter 7 · Integrals

Exercise 7.4 · Q11: ∫dx9x2+6x+5\int\frac{dx}{9x^2+6x+5}

Exercise 7.4Integrals of some particular functions · Section 7.4
The question

Integrate the function (Exercise 7.4, "Integrate the functions in Exercises 1 to 23"):

19x2+6x+5\frac{1}{9x^2 + 6x + 5}
Exam answer

Let I=∫dx9x2+6x+5I = \displaystyle\int\frac{dx}{9x^2 + 6x + 5}⁠.

9x2+6x+5=9(x2+23x+59)=9[(x+13)2+(23)2]\begin{aligned} &9x^2 + 6x + 5 \\ &= 9\left(x^2 + \frac23x + \frac59\right) \\ &= 9\left[\left(x + \frac13\right)^2 + \left(\frac23\right)^2\right] \end{aligned}

Put x+13=tx + \frac13 = t⁠. Then dx=dtdx = dt⁠.

I=19∫dtt2+(23)2I = \frac19\int\frac{dt}{t^2 + \left(\frac23\right)^2}

[Using ∫dtt2+a2\displaystyle\int\frac{dt}{t^2 + a^2} =1atan⁡−1ta+C= \frac1a\tan^{-1}\frac ta + C]

I=19⋅32tan⁡−13t2+C=16tan⁡−13t2+C\begin{aligned} I &= \frac19\cdot\frac32\tan^{-1}\frac{3t}{2} + C \\ &= \frac16\tan^{-1}\frac{3t}{2} + C \end{aligned}

∴ I=16tan⁡−1(3x+12)+C\therefore\ I = \dfrac16\tan^{-1}\left(\dfrac{3x + 1}{2}\right) + C⁠.

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

The one idea this problem is built on

This is type (7) of Section 7.4, ∫dxax2+bx+c\int\frac{dx}{ax^2 + bx + c}⁠. Take out the leading coefficient 9, complete the square, and the denominator becomes 9[(x+13)2+(23)2]9\left[\left(x + \frac13\right)^2 + \left(\frac23\right)^2\right]⁠. The leftover is positive, so formula 7.4 (3) gives an inverse tangent.

Given, to find, and the plan
Given
f(x)=19x2+6x+5f(x) = \dfrac{1}{9x^2 + 6x + 5}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Take out 9 and complete the square. (2) Read off the domain. (3) Shift t=x+13t = x + \frac13 and use formula 7.4 (3) with a=23a = \frac23⁠. (4) Return to xx and simplify. (5) Differentiate back.

Basics you need

1Completing the square

For ax2+bx+cax^2 + bx + c⁠, first take aa out: a(x2+bax+ca)a\left(x^2 + \frac bax + \frac ca\right)⁠, then complete the square inside. Basics: completing the square

2Formula 7.4 (3)

∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C⁠. Lesson: integrals of particular functions

3Derivative of tan⁡−1\tan^{-1}

ddutan⁡−1u=11+u2\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2}⁠. Basics: integration undoes differentiation

Solution

  1. Take out 9 and complete the square

    9x2+6x+5=9(x2+23x+59)=9[(x+13)2−19+59]=9[(x+13)2+(23)2].\begin{aligned} 9x^2 + 6x + 5 &= 9\left(x^2 + \frac23x + \frac59\right) \\ &= 9\left[\left(x + \frac13\right)^2 - \frac19 + \frac59\right] \\ &= 9\left[\left(x + \frac13\right)^2 + \left(\frac23\right)^2\right]. \end{aligned}
    Half of 23\frac23 is 13\frac13⁠, whose square is 19\frac19⁠; adding and subtracting 19\frac19 creates (x+13)2\left(x + \frac13\right)^2⁠. The leftover is −19+59=49=(23)2-\frac19 + \frac59 = \frac49 = \left(\frac23\right)^2⁠, positive.
    Basics: completing the square

  2. Where is the integrand defined?

    By Step 1 the denominator is at least 9⋅49=4>09\cdot\frac49 = 4 > 0⁠, so it is never 0. The integrand is defined for every real xx⁠.

    A positive leftover after completing the square means the quadratic has no real roots (its discriminant 36−180=−14436 - 180 = -144 is negative).

  3. Shift t=x+13t = x + \frac13 and use formula 7.4 (3)

    Put t=x+13t = x + \frac13⁠, dt=dxdt = dx⁠:

    ∫dx9x2+6x+5=19∫dtt2+(23)2=19⋅32tan⁡−1t2/3+C=16tan⁡−13t2+C.\begin{aligned} \int\frac{dx}{9x^2 + 6x + 5} &= \frac19\int\frac{dt}{t^2 + \left(\frac23\right)^2} \\ &= \frac19\cdot\frac32\tan^{-1}\frac{t}{2/3} + C \\ &= \frac16\tan^{-1}\frac{3t}{2} + C. \end{aligned}
    Formula 7.4 (3) with a=23a = \frac23⁠: 1a=32\frac1a = \frac32⁠, and 19⋅32=16\frac19\cdot\frac32 = \frac16⁠. Dividing tt by 23\frac23 is multiplying by 32\frac32⁠.
    Formula 7.4 (3). Lesson: integrals of particular functions

  4. Return to xx

    3t2=32(x+13)=3x+12\frac{3t}{2} = \frac{3}{2}\left(x + \frac13\right) = \frac{3x + 1}{2}⁠, so

    ∫dx9x2+6x+5=16tan⁡−13x+12+C.\int\frac{dx}{9x^2 + 6x + 5} = \frac16\tan^{-1}\frac{3x + 1}{2} + C.
    A quicker route reaches the same place: 9x2+6x+5=(3x+1)2+49x^2 + 6x + 5 = (3x + 1)^2 + 4⁠, and with u=3x+1u = 3x + 1⁠, du=3 dxdu = 3\,dx⁠, the integral is 13∫duu2+22=13⋅12tan⁡−1u2\frac13\int\frac{du}{u^2 + 2^2} = \frac13\cdot\frac12\tan^{-1}\frac u2⁠, again 16tan⁡−13x+12\frac16\tan^{-1}\frac{3x + 1}{2}⁠.

  5. Check: differentiate the answer back

    Let F(x)=16tan⁡−1uF(x) = \frac16\tan^{-1}u⁠, u=3x+12u = \frac{3x + 1}{2}⁠, u′=32u' = \frac32⁠. Then

    F′(x)=16⋅11+(3x+1)24⋅32=14⋅44+(3x+1)2=19x2+6x+5. ✓\begin{aligned} F'(x) &= \frac16\cdot\frac{1}{1 + \frac{(3x + 1)^2}{4}}\cdot\frac32 \\ &= \frac14\cdot\frac{4}{4 + (3x + 1)^2} \\ &= \frac{1}{9x^2 + 6x + 5}. \ \checkmark \end{aligned}
    16⋅32=14\frac16\cdot\frac32 = \frac14⁠; multiplying top and bottom of the inner fraction by 4 clears it; and 4+(3x+1)2=4+9x2+6x+14 + (3x + 1)^2 = 4 + 9x^2 + 6x + 1⁠, which is 9x2+6x+59x^2 + 6x + 5⁠.
    Standard formula (x), Section 7.2, with the chain rule. Basics: integration undoes differentiation
Result

∫dx9x2+6x+5=16tan⁡−1(3x+12)+C,\int\frac{dx}{9x^2 + 6x + 5} = \frac16\tan^{-1}\left(\frac{3x + 1}{2}\right) + C, valid for every real xx⁠. (Textbook answer: 16tan⁡−1(3x+12)+C\frac16\tan^{-1}\left(\frac{3x + 1}{2}\right) + C⁠, the same.)

Check it with numbers

Take x=1x = 1 and h=0.001h = 0.001⁠:

  • F(1.001)=0.18457476F(1.001) = 0.18457476 and F(0.999)=0.18447476F(0.999) = 0.18447476⁠,
  • slope ≈0.18457476−0.184474760.002=0.0500\approx \frac{0.18457476 - 0.18447476}{0.002} = 0.0500⁠,
  • integrand: 19+6+5=120=0.0500\frac{1}{9 + 6 + 5} = \frac{1}{20} = 0.0500⁠. ✓
Common mistakes
  1. Completing the square without taking out the 9, for example writing (x+3)2−4(x + 3)^2 - 4⁠. The square must be of x+b2a=x+13x + \frac{b}{2a} = x + \frac13⁠.
  2. Forgetting the 19\frac19 in front after factoring.
  3. Using a=49a = \frac49 instead of a=23a = \frac23⁠: the formula needs a2=49a^2 = \frac49⁠.
What to take away
  1. Recipe for ∫dxax2+bx+c\int\frac{dx}{ax^2 + bx + c}⁠: factor out aa⁠, complete the square, shift, then 7.4 (3) if the leftover is positive, 7.4 (1) if negative.
  2. When aa is a perfect square (here 9=329 = 3^2⁠), ax2+bx+c=(3x+… )2+…ax^2 + bx + c = (3x + \dots)^2 + \dots is a quick alternative.
  3. Always simplify ta\frac{t}{a} back to a clean expression in xx⁠.

Every fact used, and where it comes from

FactUsed inSource
Take out aa⁠, complete the squareSteps 1, 2Type (7), Section 7.4 (Basics: completing the square)
Shift t=x+13t = x + \frac13⁠, dt=dxdt = dxStep 3Section 7.3.1 (Lesson: substitution)
∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + CSteps 3, 4Formula 7.4 (3), Section 7.4
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}⁠; chain ruleStep 5Standard formula (x), Section 7.2; Theorem 4, Section 5.3.1
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