Exercise 7.5 · Q15: $\int\frac{dx}{x^4 - 1}$

Basics you need

Solution

1. Factor the denominator

   $$x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1).$$

   Further, $x^2 - 1 = (x - 1)(x + 1)$⁠, while $x^2 + 1$ has no real roots (discriminant $-4 < 0$⁠) and cannot be factorised.

   Partial fractions start from the factors of the denominator. The difference of squares is used twice. The integrand is proper (degree 0 over degree 4), so no division is needed.
   Basics: splitting fractions

2. Split using $y = x^2$ (as in Example 14)

   With $y = x^2$ the integrand is $\frac{1}{(y - 1)(y + 1)}$⁠. Write

   $$\frac{1}{(y - 1)(y + 1)} = \frac{A}{y - 1} + \frac{B}{y + 1},$$

   so $1 = A(y + 1) + B(y - 1)$⁠. Put $y = 1$⁠: $1 = 2A$⁠, $A = \frac12$⁠. Put $y = -1$⁠: $1 = -2B$⁠, $B = -\frac12$⁠. Hence

   $$\frac{1}{x^4 - 1} = \frac12\cdot\frac{1}{x^2 - 1} - \frac12\cdot\frac{1}{x^2 + 1}.$$
   This is only an algebraic identity: $y$ stands for $x^2$ while we find the constants, and is replaced by $x^2$ at once. As the book remarks after Example 14, the substitution is made only for the partial fraction part, not for the integration (we still integrate with respect to $x$⁠). The cleared identity in $y$ holds for all $y$⁠, so substituting $y = \pm 1$ is allowed.
   Table 7.2, row 1; Example 14. Lesson: partial fractions

3. Integrate the two pieces

   Section 7.4, formula (1) with $a = 1$⁠, and formula (3) with $a = 1$⁠:

   $$\begin{aligned} \int\frac{dx}{x^2 - 1} &= \frac12\log\left|\frac{x - 1}{x + 1}\right| + C_1, \\ \int\frac{dx}{x^2 + 1} &= \tan^{-1}x + C_2. \end{aligned}$$

   Therefore

   $$\begin{aligned} \int\frac{dx}{x^4 - 1} &= \frac14\log\left|\frac{x - 1}{x + 1}\right| \\ &\quad - \frac12\tan^{-1}x + C. \end{aligned}$$
   Multiply each standard result by its constant $\pm\frac12$⁠: $\frac12\cdot\frac12 = \frac14$ in front of the logarithm, and $-\frac12$ in front of the inverse tangent. The constants combine into one $C$⁠.
   Section 7.4, formulae (1) and (3); Property (IV), constant multiples. Lesson: integrals of particular functions

4. Cross-check with the full decomposition

   Using all three factors (Table 7.2 style, with $Cx + D$ over the quadratic):

   $$\frac{1}{(x - 1)(x + 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}.$$

   Clearing denominators,

   $$\begin{aligned} 1 &= A(x + 1)(x^2 + 1) + B(x - 1)(x^2 + 1) \\ &\quad + (Cx + D)(x^2 - 1). \end{aligned}$$

   Put $x = 1$⁠: $1 = 4A$⁠, $A = \frac14$⁠. Put $x = -1$⁠: $1 = -4B$⁠, $B = -\frac14$⁠. Coefficient of $x^3$⁠: $A + B + C = 0$⁠, so $C = 0$⁠. Constant term: $A - B - D = 1$⁠, so $D = \frac12 - 1 = -\frac12$⁠. Hence

   $$\frac{1}{x^4 - 1} = \frac{1}{4(x - 1)} - \frac{1}{4(x + 1)} - \frac{1}{2(x^2 + 1)},$$

   which integrates to $\frac14\log|x - 1| - \frac14\log|x + 1| - \frac12\tan^{-1}x + C$⁠: the same answer.

   Table 7.2 lists the pattern "one linear factor times one quadratic"; with two linear factors and one quadratic the same principle applies (a constant over each linear factor, $Cx + D$ over the quadratic). Getting the same result two ways is a strong check on the constants.
   Section 7.5. Basics: splitting fractions

5. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\Big(\tfrac14\log|x - 1| - \tfrac14\log|x + 1| - \tfrac12\tan^{-1}x\Big) \\ &= \frac{1}{4(x - 1)} - \frac{1}{4(x + 1)} - \frac{1}{2(1 + x^2)} \\ &= \frac{2}{4(x^2 - 1)} - \frac{1}{2(x^2 + 1)} \\ &= \frac12\cdot\frac{(x^2 + 1) - (x^2 - 1)}{(x^2 - 1)(x^2 + 1)} \\ &= \frac{1}{x^4 - 1}. \ \checkmark \end{aligned}$$

   Valid for every $x \ne \pm 1$⁠.

   $\frac{1}{x - 1} - \frac{1}{x + 1} = \frac{(x + 1) - (x - 1)}{x^2 - 1} = \frac{2}{x^2 - 1}$⁠; then the two fractions over $x^2 \mp 1$ combine, and $(x^2 + 1) - (x^2 - 1) = 2$ cancels the $\frac12$⁠.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

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