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Chapter 7 · Integrals

Exercise 7.4 · Q24 (MCQ): ∫dxx2+2x+2\int\frac{dx}{x^2 + 2x + 2}

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

Choose the correct answer.

∫dxx2+2x+2  equals\int\frac{dx}{x^2 + 2x + 2} \ \text{ equals}

(A) xtan⁡−1(x+1)+Cx\tan^{-1}(x + 1) + C    (B) tan⁡−1(x+1)+C\tan^{-1}(x + 1) + C
(C) (x+1)tan⁡−1x+C(x + 1)\tan^{-1}x + C    (D) tan⁡−1x+C\tan^{-1}x + C

Exam answer

Here x2+2x+2=(x+1)2+12x^2 + 2x + 2 = (x + 1)^2 + 1^2⁠. [Using ∫dxx2+a2\displaystyle\int\frac{dx}{x^2 + a^2} =1atan⁡−1xa+C= \frac1a\tan^{-1}\frac xa + C]

∫dx(x+1)2+12=tan⁡−1(x+1)+C\begin{aligned} &\int\frac{dx}{(x + 1)^2 + 1^2} \\ &= \tan^{-1}(x + 1) + C \end{aligned}

Answer: (B) tan⁡−1(x+1)+C\tan^{-1}(x + 1) + C

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

The one idea this problem is built on

Complete the square in the denominator: x2+2x+2=(x+1)2+1x^2 + 2x + 2 = (x + 1)^2 + 1⁠. With t=x+1t = x + 1 the integral becomes ∫dtt2+1=tan⁡−1t+C\int\frac{dt}{t^2 + 1} = \tan^{-1}t + C (formula 7.4 (3) with a=1a = 1⁠). So the answer is tan⁡−1(x+1)+C\tan^{-1}(x + 1) + C⁠, option (B).

Given, to find, and the plan
Given
The integrand 1x2+2x+2\frac{1}{x^2 + 2x + 2} and four options.
To find
Which option is ∫dxx2+2x+2\int\frac{dx}{x^2 + 2x + 2}⁠.
Plan
(1) Find the domain. (2) Complete the square. (3) Shift t=x+1t = x + 1 and use 7.4 (3). (4) Differentiate back. (5) Test every option.

Basics you need

1Completing the square

x2+2x=(x+1)2−1x^2 + 2x = (x + 1)^2 - 1⁠. 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; with a=1a = 1 it is standard formula (x), ∫dx1+x2=tan⁡−1x+C\int\frac{dx}{1 + x^2} = \tan^{-1}x + C⁠. Lesson: integrals of particular functions

3Testing an option

An option F+CF + C is right exactly when F′F' equals the integrand at every point. Basics: integration undoes differentiation

Solution

  1. The domain

    x2+2x+2=(x+1)2+1≥1x^2 + 2x + 2 = (x + 1)^2 + 1 \ge 1⁠, so the denominator is never 0 and the integrand is defined for every real xx⁠.

    A square is never negative. (Equivalently, the discriminant 4−8=−44 - 8 = -4 is negative.)

  2. Complete the square

    x2+2x+2=(x2+2x+1)+1=(x+1)2+12.x^2 + 2x + 2 = \left(x^2 + 2x + 1\right) + 1 = (x + 1)^2 + 1^2.
    Half of the coefficient 2 is 1, and 12=11^2 = 1⁠; the leftover 2−1=12 - 1 = 1 is positive, so the inverse tangent formula applies.
    Type 7.4 (7). Basics: completing the square

  3. Shift t=x+1t = x + 1 and integrate

    Put t=x+1t = x + 1⁠, dt=dxdt = dx⁠. By 7.4 (3) with a=1a = 1⁠:

    ∫dxx2+2x+2=∫dtt2+1=tan⁡−1t+C=tan⁡−1(x+1)+C.\begin{aligned} \int\frac{dx}{x^2 + 2x + 2} &= \int\frac{dt}{t^2 + 1} = \tan^{-1}t + C \\ &= \tan^{-1}(x + 1) + C. \end{aligned}
    The shift does not change dxdx⁠, since ddx(x+1)=1\frac{d}{dx}(x + 1) = 1⁠. With a=1a = 1 the formula's factor 1a\frac1a and the ta\frac ta inside are just 1 and tt⁠.
    Formula 7.4 (3); substitution, Section 7.3.1. Lesson: integrals of particular functions

  4. Check: differentiate the answer

    ddxtan⁡−1(x+1)=11+(x+1)2=1x2+2x+2. ✓\frac{d}{dx}\tan^{-1}(x + 1) = \frac{1}{1 + (x + 1)^2} = \frac{1}{x^2 + 2x + 2}. \ \checkmark
    Chain rule with inner function x+1x + 1⁠, whose derivative is 1; then 1+(x+1)2=x2+2x+21 + (x + 1)^2 = x^2 + 2x + 2 by Step 2.
    Derivative (x), Section 7.2; chain rule (Theorem 4, Section 5.3.1). Basics: integration undoes differentiation

  5. Examine every option

    An option is right exactly when its derivative is the integrand. Test at x=0x = 0⁠, where the integrand is 12\frac12⁠.

    • (A) By the product rule, ddx[xtan⁡−1(x+1)]\frac{d}{dx}\left[x\tan^{-1}(x + 1)\right] =tan⁡−1(x+1)+xx2+2x+2= \tan^{-1}(x + 1) + \frac{x}{x^2 + 2x + 2}⁠. At x=0x = 0⁠: tan⁡−11+0=π4≈0.785≠12\tan^{-1}1 + 0 = \frac{\pi}{4} \approx 0.785 \ne \frac12⁠. Wrong.
    • (B) Derivative 1x2+2x+2\frac{1}{x^2 + 2x + 2} for every xx (Step 4). At x=0x = 0⁠: 12\frac12⁠. ✓ Correct.
    • (C) ddx[(x+1)tan⁡−1x]=tan⁡−1x+x+11+x2\frac{d}{dx}\left[(x + 1)\tan^{-1}x\right] = \tan^{-1}x + \frac{x + 1}{1 + x^2}⁠. At x=0x = 0⁠: 0+1=1≠120 + 1 = 1 \ne \frac12⁠. Wrong.
    • (D) ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}⁠. At x=0x = 0⁠: 1≠121 \ne \frac12⁠. Wrong (it forgets the shift x+1x + 1⁠).
    A single point where the derivatives differ is enough to reject an option. Options (A) and (C) wrongly treat the integral like "multiply by xx" or "multiply by x+1x + 1"; the product rule shows they produce extra terms.
    Product rule, Section 5.3 (Chapter 5). Basics: integration undoes differentiation
Result ∫dxx2+2x+2=tan⁡−1(x+1)+C,\int\frac{dx}{x^2 + 2x + 2} = \tan^{-1}(x + 1) + C,

valid for every real xx⁠. The correct option is (B). (Textbook answer: B.)

Check it with numbers

By Section 7.8, ∫01dxx2+2x+2=F(1)−F(0)\int_0^1 \frac{dx}{x^2 + 2x + 2} = F(1) - F(0) with F(x)=tan⁡−1(x+1)F(x) = \tan^{-1}(x + 1)⁠:

  • a computer's sum of very many thin strips (found without our answer) gives 0.32175060.3217506⁠;
  • F(1)−F(0)=tan⁡−12−tan⁡−11F(1) - F(0) = \tan^{-1}2 - \tan^{-1}1 =1.10714872−0.78539816= 1.10714872 - 0.78539816 =0.32175056= 0.32175056⁠, which rounds to 0.32175060.3217506⁠. ✓
Common mistakes
  1. Choosing (D). Completing the square shifts the variable: the answer contains x+1x + 1⁠, not xx⁠.
  2. Choosing (A) or (C). These multiply an inverse tangent by xx or x+1x + 1⁠; differentiating with the product rule shows extra terms that are not in the integrand.
  3. Using the log formula. The leftover after completing the square is +1+1⁠, so it is tan⁡−1\tan^{-1}⁠, not 7.4 (1).
What to take away
  1. ∫dx(x+h)2+a2=1atan⁡−1x+ha+C\int\frac{dx}{(x + h)^2 + a^2} = \frac1a\tan^{-1}\frac{x + h}{a} + C⁠.
  2. For MCQs, differentiate each option and compare with the integrand at a convenient point.

Every fact used, and where it comes from

FactUsed inSource
Completing the squareSteps 1, 2Type 7.4 (7) (Basics: completing the square)
∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + CStep 3Formula 7.4 (3)
Substitution t=x+1t = x + 1Step 3Section 7.3.1 (Basics: substitution)
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}⁠; chain ruleSteps 4, 5Derivative (x), Section 7.2; Theorem 4, Section 5.3.1
Product ruleStep 5Section 5.3 (Chapter 5)
∫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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