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Chapter 7 · Integrals

Exercise 7.4 · Q10: ∫dxx2+2x+2\int\frac{dx}{\sqrt{x^2+2x+2}}

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"):

1x2+2x+2\frac{1}{\sqrt{x^2 + 2x + 2}}
Exam answer

Let I=∫dxx2+2x+2I = \displaystyle\int\frac{dx}{\sqrt{x^2 + 2x + 2}}⁠.

x2+2x+2=(x+1)2+12x^2 + 2x + 2 = (x + 1)^2 + 1^2

Put x+1=tx + 1 = t⁠. Then dx=dtdx = dt⁠.

I=∫dtt2+12I = \int\frac{dt}{\sqrt{t^2 + 1^2}}

[Using ∫dtt2+a2\displaystyle\int\frac{dt}{\sqrt{t^2 + a^2}} =log⁡∣t+t2+a2∣+C= \log\left|t + \sqrt{t^2 + a^2}\right| + C]

I=log⁡∣t+t2+1∣+CI = \log\left|t + \sqrt{t^2 + 1}\right| + C

∴ I=log⁡∣x+1+x2+2x+2∣\therefore\ I = \log\left|x + 1 + \sqrt{x^2 + 2x + 2}\right| +C+ C⁠.

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

The one idea this problem is built on

Complete the square: x2+2x+2=(x+1)2+1x^2 + 2x + 2 = (x + 1)^2 + 1⁠. After the shift t=x+1t = x + 1 the integral is ∫dtt2+1\int\frac{dt}{\sqrt{t^2 + 1}}⁠, the standard form 7.4 (6). This is type (8) of Section 7.4: ∫dxax2+bx+c\int\frac{dx}{\sqrt{ax^2 + bx + c}}⁠.

Given, to find, and the plan
Given
f(x)=1x2+2x+2f(x) = \dfrac{1}{\sqrt{x^2 + 2x + 2}}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Complete the square. (2) Use it to find the domain. (3) Shift t=x+1t = x + 1 and use formula 7.4 (6). (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Completing the square

x2+px=(x+p2)2−p24x^2 + px = \left(x + \frac p2\right)^2 - \frac{p^2}{4}⁠. Basics: completing the square

2A shift is a substitution

t=x+1t = x + 1 gives dt=dxdt = dx⁠: nothing else changes. Basics: substitution (the chain rule backwards)

3Formula 7.4 (6)

∫dxx2+a2=log⁡∣x+x2+a2∣+C\int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C⁠. Lesson: integrals of particular functions

Solution

  1. Complete the square

    x2+2x+2=x2+2x+1−1+2=(x+1)2+1.\begin{aligned} x^2 + 2x + 2 &= x^2 + 2x + 1 - 1 + 2 \\ &= (x + 1)^2 + 1. \end{aligned}
    Half the coefficient of xx is 1, and 12=11^2 = 1⁠. Adding and subtracting 1 creates the perfect square x2+2x+1=(x+1)2x^2 + 2x + 1 = (x + 1)^2 without changing the value. What is left over, +1+1⁠, is positive.
    Basics: completing the square

  2. Where is the integrand defined?

    By Step 1, x2+2x+2=(x+1)2+1≥1>0x^2 + 2x + 2 = (x + 1)^2 + 1 \ge 1 > 0 for every real xx⁠. So the root is always of a positive number and is never 0: the integrand is defined on the whole real line.

    Completing the square shows the sign at a glance. (Equivalently, the discriminant 22−4⋅2=−42^2 - 4\cdot 2 = -4 is negative, so the quadratic has no real roots and keeps the sign of its leading coefficient.)

  3. Shift t=x+1t = x + 1 and use formula 7.4 (6)

    Put t=x+1t = x + 1⁠, dt=dxdt = dx⁠:

    ∫dx(x+1)2+1=∫dtt2+12=log⁡∣t+t2+1∣+C.\begin{aligned} \int\frac{dx}{\sqrt{(x + 1)^2 + 1}} &= \int\frac{dt}{\sqrt{t^2 + 1^2}} \\ &= \log\left|t + \sqrt{t^2 + 1}\right| + C. \end{aligned}
    The positive leftover makes the shape "square plus constant, under a root", which is formula 7.4 (6) with a=1a = 1⁠. If the leftover had been negative, we would have used 7.4 (4) instead.
    Formula 7.4 (6); type (8), Section 7.4. Lesson: integrals of particular functions

  4. Return to xx

    With t=x+1t = x + 1⁠, t2+1=x2+2x+2t^2 + 1 = x^2 + 2x + 2 (Step 1 read backwards):

    ∫dxx2+2x+2=log⁡∣x+1+x2+2x+2∣+C.\begin{aligned} &\int\frac{dx}{\sqrt{x^2 + 2x + 2}} \\ &= \log\left|x + 1 + \sqrt{x^2 + 2x + 2}\right| + C. \end{aligned}
    Writing the root in terms of the original quadratic gives the textbook's form. The inside of the modulus is positive for every xx⁠, since (x+1)2+1>∣x+1∣\sqrt{(x + 1)^2 + 1} > |x + 1|⁠.

  5. Check: differentiate the answer back

    Let s=x2+2x+2s = \sqrt{x^2 + 2x + 2}⁠, so s′=2x+22s=x+1ss' = \frac{2x + 2}{2s} = \frac{x + 1}{s}⁠. With F(x)=log⁡(x+1+s)F(x) = \log(x + 1 + s)⁠,

    F′(x)=1+x+1sx+1+s=s+x+1s (x+1+s)=1s.\begin{aligned} F'(x) &= \frac{1 + \frac{x + 1}{s}}{x + 1 + s} \\ &= \frac{s + x + 1}{s\,(x + 1 + s)} = \frac1s. \end{aligned}

    So F′(x)=1x2+2x+2F'(x) = \frac{1}{\sqrt{x^2 + 2x + 2}}⁠, the integrand. ✓\checkmark

    Chain rule for log⁡u\log u and for v\sqrt v⁠. The sum x+1+sx + 1 + s appears on top and bottom and cancels.
    Basics: integration undoes differentiation
Result

∫dxx2+2x+2=log⁡∣x+1+x2+2x+2∣+C,\begin{aligned} &\int\frac{dx}{\sqrt{x^2 + 2x + 2}} \\ &= \log\left|x + 1 + \sqrt{x^2 + 2x + 2}\right| + C, \end{aligned} valid for every real xx⁠. (Textbook answer: log⁡∣x+1+x2+2x+2∣+C\log\left|x + 1 + \sqrt{x^2 + 2x + 2}\right| + C⁠, the same.)

Check it with numbers

Take x=1x = 1 and h=0.001h = 0.001⁠:

  • F(1.001)=1.44408260F(1.001) = 1.44408260 and F(0.999)=1.44318817F(0.999) = 1.44318817⁠,
  • slope ≈1.44408260−1.443188170.002=0.4472\approx \frac{1.44408260 - 1.44318817}{0.002} = 0.4472⁠,
  • integrand: 11+2+2=15=0.4472\frac{1}{\sqrt{1 + 2 + 2}} = \frac{1}{\sqrt5} = 0.4472⁠. ✓
Common mistakes
  1. Completing the square wrongly, for example (x+2)2−2(x + 2)^2 - 2⁠. Half of the coefficient 2 is 1, not 2.
  2. Answering log⁡∣x+x2+2x+2∣\log\left|x + \sqrt{x^2 + 2x + 2}\right|⁠. The term outside the root is the shifted variable t=x+1t = x + 1⁠, not xx⁠.
  3. Using sin⁡−1\sin^{-1} because there is a root. sin⁡−1\sin^{-1} needs "constant minus square" under the root.
What to take away
  1. For 1quadratic\frac{1}{\sqrt{\text{quadratic}}}⁠: complete the square, shift, then choose among 7.4 (4), (5), (6) by the signs.
  2. ∫dx(x+h)2+k2\int\frac{dx}{\sqrt{(x + h)^2 + k^2}} equals log⁡∣x+h+(x+h)2+k2∣+C\log\left|x + h + \sqrt{(x + h)^2 + k^2}\right| + C⁠.
  3. Completing the square also tells you the domain.

Every fact used, and where it comes from

FactUsed inSource
Completing the squareSteps 1, 2Type (8), Section 7.4 (Basics: completing the square)
Shift t=x+1t = x + 1⁠, dt=dxdt = dxStep 3Section 7.3.1 (Lesson: substitution)
∫dxx2+a2=log⁡∣x+x2+a2∣+C\int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + CStep 3Formula 7.4 (6), Section 7.4
Chain rule; F′=fF' = f checks the answerStep 5Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
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