Exercise 7.5 · Q2: $\int\frac{1}{x^2-9}\,dx$

Basics you need

Solution

1. Factor the denominator; the fraction is proper

   $$x^2 - 9 = x^2 - 3^2 = (x-3)(x+3).$$

   The numerator 1 has degree 0 and the denominator has degree 2, so the fraction is proper. The two linear factors are different ($a = 3$⁠, $b = -3$⁠).

   Table 7.2 needs the denominator written as a product of factors, and a proper fraction. Both conditions are now visible.
   Section 7.5. Basics: splitting fractions

2. Write the form of the decomposition

   $$\frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3} \qquad (1)$$
   One unknown constant over each different linear factor, exactly as in row 1 of Table 7.2 (here $p = 0$⁠, $q = 1$⁠).
   Table 7.2, row 1. Lesson: partial fractions

3. Clear the denominators

   Multiply (1) by $(x-3)(x+3)$⁠:

   $$1 = A(x+3) + B(x-3). \qquad (2)$$
   (1) is an identity for $x \ne \pm 3$⁠, so the polynomials in (2) agree at infinitely many points and are therefore equal for every $x$⁠, including $x = \pm 3$⁠.
   Remark after Example 11, Section 7.5.

4. Find $A$ and $B$

   Put $x = 3$ in (2): $1 = A(6) + 0$⁠, so $A = \frac16$⁠.

   Put $x = -3$ in (2): $1 = 0 + B(-6)$⁠, so $B = -\frac16$⁠.

   Confirm by comparing coefficients. The right side of (2) is $(A+B)x + (3A - 3B)$⁠, and the left side is $0\cdot x + 1$⁠:

   $$\begin{aligned} \text{coefficient of } x:&\quad A + B = \tfrac16 - \tfrac16 = 0, \\ \text{constant term}:&\quad 3A - 3B = \tfrac12 + \tfrac12 = 1. \end{aligned}$$
   Putting in a root kills one bracket and isolates the other constant. Comparing coefficients is the book's method; doing both catches arithmetic slips.
   Basics: splitting fractions

5. Integrate the two simple fractions

   $$\frac{1}{x^2-9} = \frac16\cdot\frac{1}{x-3} - \frac16\cdot\frac{1}{x+3},$$ $$\begin{aligned} \int\frac{dx}{x^2-9} &= \frac16\int\frac{dx}{x-3} - \frac16\int\frac{dx}{x+3} \\ &= \frac16\log|x-3| - \frac16\log|x+3| + C \\ &= \frac16\log\left|\frac{x-3}{x+3}\right| + C. \end{aligned}$$
   Integral of a sum, constants outside (Properties (III), (IV)); each piece is $\int\frac{dx}{x+k} = \log|x+k|$⁠. Finally $\log|p| - \log|q| = \log\left|\frac pq\right|$⁠.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

6. Cross-check with formula (1) of Section 7.4

   With $a = 3$⁠, formula (1) gives

   $$\int\frac{dx}{x^2-3^2} = \frac{1}{2\cdot 3}\log\left|\frac{x-3}{x+3}\right| + C,$$

   which is exactly our answer, since $\frac{1}{2\cdot 3} = \frac16$⁠.

   The book proves formula (1) by this very decomposition. Seeing the two routes agree confirms both the constants and the order of the quotient $\frac{x-3}{x+3}$⁠.
   Section 7.4, formula (1). Lesson: integrals of particular functions

7. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\Big(\tfrac16\log|x-3| - \tfrac16\log|x+3|\Big) \\ &= \frac16\left(\frac{1}{x-3} - \frac{1}{x+3}\right) \\ &= \frac16\cdot\frac{(x+3) - (x-3)}{(x-3)(x+3)} \\ &= \frac16\cdot\frac{6}{x^2-9} = \frac{1}{x^2-9}. \ \checkmark \end{aligned}$$
   The derivative returns the integrand, so the antiderivative is right. It is valid on each interval avoiding $x = 3$ and $x = -3$⁠.
   $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2, list item (xii). Basics: integration undoes differentiation

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