Chapter 7 · Integrals

Exercise 7.8 · Q11: $\int_2^3\frac{dx}{x^2 - 1}$

Exercise 7.8Fundamental theorem of calculus · Section 7.8

The question

Evaluate the definite integral

\int_2^3\frac{dx}{x^2 - 1}.

Exam answer

Let $I = \displaystyle\int_2^3\frac{dx}{x^2 - 1}$ ⁠.

Using $\displaystyle\int\frac{dx}{x^2 - a^2}$ $= \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C,$

\begin{aligned} I &= \frac12\left[\log\left|\frac{x - 1}{x + 1}\right|\right]_2^3 \\ &= \frac12\left(\log\frac24 - \log\frac13\right) \\ &= \frac12\log\left(\frac12\cdot\frac31\right) \end{aligned}

$\therefore\ I = \dfrac12\log\dfrac32$ ⁠.

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

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3): $\int_a^b f(x)\,dx = F(b) - F(a)$ ⁠. The denominator factorises, $x^2 - 1 = (x - 1)(x + 1)$ ⁠, and the fraction splits into $\frac12\left(\frac{1}{x-1} - \frac{1}{x+1}\right)$ ⁠. Each piece integrates to a logarithm, which is formula (1) of Section 7.4 with $a = 1$ ⁠: $\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C$ ⁠.

Given, to find, and the plan

Given: $f(x) = \frac{1}{x^2 - 1}$ on $[2, 3]$ ⁠.
To find: The exact value of $\int_2^3\frac{dx}{x^2 - 1}$ ⁠.
Plan: (1) Continuity: the denominator vanishes only at $\pm 1$ ⁠. (2) Split the fraction and integrate; check by differentiating. (3) $F(3) - F(2)$ with the laws of logarithms. (4) Size check against the graph.

Basics you need

1Second fundamental theorem

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$ for $f$ continuous on $[a, b]$ and $F' = f$ ⁠. Lesson: definite integrals and the fundamental theorem

2Splitting a fraction

$\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ (partial fractions, Section 7.5; the same split proves formula (1) of Section 7.4). Basics: splitting fractions

3Logs

$\int\frac{dx}{x + k} = \log|x + k| + C$ (standard formula (xii) with a shift); $\log a - \log b = \log\frac{a}{b}$ ⁠. Basics: logarithms and 1/x

Solution

Check continuity on $[2, 3]$

$x^2 - 1 = (x - 1)(x + 1) = 0$ only at $x = 1$ and $x = -1$ ⁠. Neither lies in $[2, 3]$ ⁠; in fact for $2 \le x \le 3$ ⁠, $x^2 - 1 \ge 3 > 0$ ⁠. So $f$ is continuous on $[2, 3]$ ⁠.

A quotient of continuous functions is continuous wherever the denominator is not 0, and Theorem 2 needs exactly this on the closed interval (Remark (iv) after Theorem 2). The graph has a vertical asymptote at $x = 1$ ⁠, well to the left of our interval.

Quotient of continuous functions: Chapter 5, Theorem 1, Section 5.2.1. Lesson: the fundamental theorem
Split the fraction

Write $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ ⁠. Multiplying by $(x-1)(x+1)$ ⁠:
$1 = A(x + 1) + B(x - 1) \quad\text{for every } x.$
Put $x = 1$ ⁠: $1 = 2A$ ⁠, so $A = \frac12$ ⁠. Put $x = -1$ ⁠: $1 = -2B$ ⁠, so $B = -\frac12$ ⁠. Hence
$\frac{1}{x^2 - 1} = \frac12\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right).$
Each simple fraction has a known integral; the product in the denominator does not. Check: $\frac{(x+1) - (x-1)}{2(x-1)(x+1)} = \frac{2}{2(x^2 - 1)} = \frac{1}{x^2 - 1}$ ⁠.

Partial fractions, Section 7.5 (Table 7.2, form 1). Lesson: partial fractions
Integrate
$\begin{aligned}\int\frac{dx}{x^2 - 1} &= \frac12\Big(\log|x - 1| - \log|x + 1|\Big) + C \\ &= \frac12\log\left|\frac{x - 1}{x + 1}\right| + C.\end{aligned}$
On $[2, 3]$ both $x - 1$ and $x + 1$ are positive, so take
$F(x) = \frac12\log\frac{x - 1}{x + 1}.$
This is formula (1) of Section 7.4, $\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C$ ⁠, with $a = 1$ ⁠. The bars can go because $\frac{x - 1}{x + 1} > 0$ for $x > 1$ ⁠.

Formula 7.4 (1); standard formula (xii). Lesson: integrals of particular functions
Check: differentiate $F$ back

For $x > 1$ ⁠, $F(x) = \frac12\left[\log(x - 1) - \log(x + 1)\right]$ ⁠, so
$\begin{aligned}F'(x) &= \frac12\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right) \\ &= \frac12\cdot\frac{(x + 1) - (x - 1)}{x^2 - 1} = \frac{1}{x^2 - 1}.\end{aligned}$
The derivative returns exactly the integrand, so $F$ is a valid antiderivative on $[2, 3]$ ⁠.

$\frac{d}{dx}\log(x + k) = \frac{1}{x + k}$ for $x + k > 0$ ⁠. Basics: integration undoes differentiation
Evaluate $F(3) - F(2)$

$F(3) = \frac12\log\frac{2}{4} = \frac12\log\frac12$ and $F(2) = \frac12\log\frac{1}{3}$ ⁠. So
$\begin{aligned}\int_2^3\frac{dx}{x^2 - 1} &= \frac12\left(\log\frac12 - \log\frac13\right) \\ &= \frac12\log\left(\frac12\cdot\frac31\right) = \frac12\log\frac32 \\ &\approx 0.2027326.\end{aligned}$
Upper minus lower, then $\log a - \log b = \log\frac{a}{b}$ ⁠: dividing $\frac12$ by $\frac13$ gives $\frac32$ ⁠. Both $\log\frac12$ and $\log\frac13$ are negative, but the larger one comes first, so the result is positive, as it must be for a positive integrand.

Laws of logarithms. Basics: logarithms
Does the size make sense?

On $[2, 3]$ ⁠, $f$ falls from $f(2) = \frac13$ to $f(3) = \frac18$ ⁠. On a base of width 1:
$\frac18 = 0.125 < 0.2027 < \frac13 \approx 0.333.$

$y = \frac{1}{x^2 - 1}$ for $x > 1$ ⁠, with the dashed asymptote $x = 1$ where it is undefined. The shaded region from 2 to 3 lies well away from it and has area $\frac12\log\frac32 \approx 0.203$ ⁠.

Since $f > 0$ on $[2, 3]$ ⁠, the integral is the shaded area (Section 7.8.1), and it lies between the lowest and highest rectangles on that base.

Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips

Result

\int_2^3\frac{dx}{x^2 - 1} = \frac12\log\frac32 \approx 0.2027326.

(Textbook answer: $\frac12\log\frac32$ ⁠, the same.)

Check it with numbers

Numerical integration of $\frac{1}{x^2 - 1}$ from 2 to 3 (thin strips, no antiderivative used) gives $0.2027326$ ⁠, and $\frac12\log 1.5$ ⁠, half of $\log 1.5 \approx 0.40546511$ ⁠, is $\approx 0.2027326$ ⁠. They agree.

Common mistakes

Forgetting the $\frac{1}{2a}$ ⁠. With $a = 1$ it is $\frac12$ ⁠; leaving it out doubles the answer.
Upside-down fraction. It is $\log\left|\frac{x - a}{x + a}\right|$ for $\frac{1}{x^2 - a^2}$ ⁠; the form $\log\left|\frac{a + x}{a - x}\right|$ belongs to $\frac{1}{a^2 - x^2}$ ⁠.
Using $\tan^{-1}$ ⁠. That is for $x^2 + a^2$ ⁠; a minus sign in the denominator gives logarithms.

What to take away

$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C$ ⁠, for $x \ne \pm a$ ⁠.
Check that the interval avoids the zeros of the denominator before using Theorem 2.
Combine the end values with the laws of logarithms into one log.

Every fact used, and where it comes from

Fact	Used in	Source
$\int_a^b f = F(b) - F(a)$ for continuous $f$	Steps 1, 5	Theorem 2, Section 7.8.3 (Lesson)
Quotient of continuous functions is continuous	Step 1	Chapter 5, Theorem 1, Section 5.2.1
Splitting into simple fractions	Step 2	Section 7.5, Table 7.2 (Lesson)
$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left\|\frac{x - a}{x + a}\right\| + C$	Step 3	Formula (1), Section 7.4 (Lesson)
$\int\frac{dx}{x} = \log\|x\| + C$	Steps 3, 4	Standard formula (xii), Section 7.2 (Basics)
$\log a - \log b = \log\frac{a}{b}$	Step 5	Laws of logarithms (Basics)
Definite integral as area	Step 6	Section 7.8.1 (Basics)

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Exercise 7.8 · Q11: ∫23dxx2−1\int_2^3\frac{dx}{x^2 - 1}∫23​x2−1dx​

Basics you need

Solution

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

Split the fraction

Integrate

Check: differentiate FFF back

Evaluate F(3)−F(2)F(3) - F(2)F(3)−F(2)

Does the size make sense?

Every fact used, and where it comes from

Exercise 7.8 · Q11: $\int_2^3\frac{dx}{x^2 - 1}$

Check continuity on $[2, 3]$

Check: differentiate $F$ back

Evaluate $F(3) - F(2)$