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Chapter 7 · Integrals

Exercise 7.8 · Q13: ∫23x dxx2+1\int_2^3 \frac{x\,dx}{x^2 + 1}

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫23x dxx2+1.\int_2^3 \frac{x\,dx}{x^2 + 1}.
Exam answer

Let I=∫23x dxx2+1I = \displaystyle\int_2^3 \frac{x\,dx}{x^2 + 1}⁠.

Put x2+1=tx^2 + 1 = t⁠. Then 2x dx=dt2x\,dx = dt⁠, so x dx=12 dtx\,dx = \frac12\,dt⁠.

When x=2x = 2⁠, t=5t = 5⁠; when x=3x = 3⁠, t=10t = 10⁠.

I=12∫510dtt=12[log⁡t]510=12(log⁡10−log⁡5)=12log⁡105.\begin{aligned} I &= \frac12\int_5^{10}\frac{dt}{t} = \frac12\Big[\log t\Big]_5^{10} \\ &= \frac12\left(\log 10 - \log 5\right) = \frac12\log\frac{10}{5}. \end{aligned}

∴ I=12log⁡2\therefore\ I = \dfrac12\log 2⁠.

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

The one idea this problem is built on

The numerator xx is half the derivative of the denominator x2+1x^2 + 1⁠. So the substitution t=x2+1t = x^2 + 1 turns the integrand into 12⋅1t\frac12\cdot\frac1t⁠, whose antiderivative is 12log⁡t\frac12\log t⁠. Then Theorem 2 (Section 7.8.3) gives F(3)−F(2)F(3) - F(2)⁠.

Given, to find, and the plan
Given
The integrand f(x)=xx2+1f(x) = \frac{x}{x^2 + 1} on [2,3][2, 3]⁠.
To find
The exact value of ∫23x dxx2+1\int_2^3\frac{x\,dx}{x^2 + 1}⁠, and a decimal.
Plan
(1) Check continuity. (2) Substitute t=x2+1t = x^2 + 1 to find FF⁠. (3) Differentiate FF back. (4) Compute F(3)−F(2)F(3) - F(2) and simplify with the laws of logarithms.

Basics you need

1Second fundamental theorem

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) when ff is continuous on [a,b][a, b] and F′=fF' = f⁠. Lesson: definite integrals and the fundamental theorem

2Substitution

If t=g(x)t = g(x) then dt=g′(x) dxdt = g'(x)\,dx⁠; rewrite the whole integrand in tt⁠. Basics: substitution (the chain rule backwards)

3Logarithm of 1t\frac1t

∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + C⁠, and log⁡a−log⁡b=log⁡ab\log a - \log b = \log\frac{a}{b}⁠. In this book "log" means the natural logarithm. Basics: logarithms and 1/x

Solution

  1. Check continuity on [2,3][2, 3]

    The numerator xx and the denominator x2+1x^2 + 1 are polynomials, and x2+1≥1>0x^2 + 1 \ge 1 > 0 for every real xx⁠. So the quotient is continuous on all of R\mathbb{R}⁠, in particular on [2,3][2, 3]⁠.

    A quotient of continuous functions is continuous wherever the denominator is not 0. Here the denominator is never 0, so Theorem 2 applies.
    x2≥0x^2 \ge 0 for every real xx⁠, so x2+1≥1x^2 + 1 \ge 1⁠. Lesson: the fundamental theorem

  2. Substitute t=x2+1t = x^2 + 1

    Then dtdx=2x\frac{dt}{dx} = 2x⁠, so dt=2x dxdt = 2x\,dx and x dx=12 dtx\,dx = \frac12\,dt⁠. Hence

    ∫x dxx2+1=12∫dtt=12log⁡∣t∣+C.\int\frac{x\,dx}{x^2 + 1} = \frac12\int\frac{dt}{t} = \frac12\log|t| + C.

    Since t=x2+1>0t = x^2 + 1 > 0⁠, ∣t∣=t|t| = t⁠, and

    ∫x dxx2+1=12log⁡(x2+1)+C.\int\frac{x\,dx}{x^2 + 1} = \frac12\log(x^2 + 1) + C.

    Take F(x)=12log⁡(x2+1)F(x) = \frac12\log(x^2 + 1)⁠.

    The derivative of the denominator, 2x2x⁠, is a constant multiple of the numerator xx⁠. That is the signal for this substitution: after it, no xx is left, only 1t\frac1t⁠.
    Standard formula (xii), Section 7.2: ∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + C⁠. Lesson: integration by substitution

  3. Check: differentiate FF back

    F′(x)=12⋅1x2+1⋅2x=xx2+1. ✓F'(x) = \frac12\cdot\frac{1}{x^2 + 1}\cdot 2x = \frac{x}{x^2 + 1}.\ \checkmark
    Chain rule with outer function 12log⁡u\frac12\log u and inner function u=x2+1u = x^2 + 1⁠. The 2 from the inner derivative cancels the 12\frac12⁠.
    ddulog⁡u=1u\frac{d}{du}\log u = \frac1u for u>0u > 0⁠. Basics: integration undoes differentiation

  4. Evaluate F(3)−F(2)F(3) - F(2)

    At x=3x = 3⁠: x2+1=10x^2 + 1 = 10⁠. At x=2x = 2⁠: x2+1=5x^2 + 1 = 5⁠. So

    I=12log⁡10−12log⁡5=12log⁡105=12log⁡2≈0.3465736.\begin{aligned}I &= \frac12\log 10 - \frac12\log 5 \\ &= \frac12\log\frac{10}{5} = \frac12\log 2 \approx 0.3465736.\end{aligned}

    With new limits instead: x=2x = 2 gives t=5t = 5⁠, x=3x = 3 gives t=10t = 10⁠, and 12∫510dtt=12(log⁡10−log⁡5)\frac12\int_5^{10}\frac{dt}{t} = \frac12(\log 10 - \log 5)⁠, the same number.

    1 2 3 0.2 0.4 x y y = x/(x² + 1) area ≈ 0.347
    The shaded region under y=xx2+1y = \frac{x}{x^2 + 1} from 2 to 3 has area 12log⁡2≈0.347\frac12\log 2 \approx 0.347⁠. The curve falls from 0.40.4 to 0.30.3 there, so the area must lie between 0.30.3 and 0.40.4⁠, and it does.
    Upper limit first, then subtract the lower. The quotient law log⁡a−log⁡b=log⁡ab\log a - \log b = \log\frac{a}{b} combines the two logs. The new limits are legitimate because t=x2+1t = x^2 + 1 is increasing on [2,3][2, 3]⁠, so tt runs once from 5 to 10 (Section 7.9).
    105=2\frac{10}{5} = 2⁠. Basics: logarithms and 1/x
Result ∫23x dxx2+1=12log⁡2≈0.3465736.\int_2^3\frac{x\,dx}{x^2 + 1} = \frac12\log 2 \approx 0.3465736.

(Textbook answer: 12log⁡2\frac12\log 2⁠.)

Check it with numbers

A numerical integration of xx2+1\frac{x}{x^2 + 1} over [2,3][2, 3] gives about 0.3465736, and 12log⁡2≈0.3465736\frac12\log 2 \approx 0.3465736⁠. They agree.

The separate values: 12log⁡10≈1.15129255\frac12\log 10 \approx 1.15129255 and 12log⁡5≈0.80471896\frac12\log 5 \approx 0.80471896⁠; their difference is 0.34657359≈0.34657360.34657359 \approx 0.3465736⁠.

Common mistakes
  1. Dropping the 12\frac12⁠. dt=2x dxdt = 2x\,dx⁠, so x dx=12dtx\,dx = \frac12 dt⁠; without the 12\frac12 the answer would be log⁡2\log 2⁠, twice too big.
  2. Writing log⁡10−log⁡5=log⁡5\log 10 - \log 5 = \log 5⁠. A difference of logs is the log of the quotient, not of the difference.
  3. Mixing limits. If you stay in tt⁠, the limits are 5 and 10, not 2 and 3.
What to take away
  1. When the numerator is a constant times the derivative of the denominator, the integral is that constant times the log of the denominator.
  2. x2+1>0x^2 + 1 > 0 always, so no absolute value bars are needed in log⁡(x2+1)\log(x^2 + 1)⁠.
  3. A quick bound (smallest height × width, largest height × width) catches arithmetic slips.

Every fact used, and where it comes from

FactUsed inSource
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) for continuous ffSteps 1, 4Theorem 2, Section 7.8.3 (Lesson)
Substitution t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dxStep 2Section 7.3.1 (Lesson)
∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 2Standard formula (xii), Section 7.2 (Basics)
Chain rule; ddulog⁡u=1u\frac{d}{du}\log u = \frac1uStep 3Chapter 5 (Basics)
log⁡a−log⁡b=log⁡ab\log a - \log b = \log\frac{a}{b}Step 4Laws of logarithms (Basics)
Changing the limits after a substitutionStep 4Section 7.9
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