Exercise 7.5 · Q20: $\int\frac{dx}{x(x^4 - 1)}$

Basics you need

Solution

1. Multiply numerator and denominator by $x^3$

   $$\frac{1}{x(x^4 - 1)} = \frac{x^3}{x^4(x^4 - 1)}.$$
   $x \ne 0$⁠, so $x^3 \ne 0$ and the value is unchanged. Now the numerator is $\frac14$ of the derivative of $x^4$⁠, and the denominator is written in terms of $x^4$ only.
   $x\cdot x^3 = x^4$⁠. Basics: powers of x

2. Substitute $t = x^4$

   Then $dt = 4x^3\,dx$⁠, so $x^3\,dx = \frac14\,dt$⁠, and

   $$\int\frac{x^3\,dx}{x^4(x^4 - 1)} = \frac14\int\frac{dt}{t(t - 1)},$$

   where $t > 0$ and $t \ne 1$ on the domain.

   The factor $x^3\,dx$ is supplied by $dt$ up to the constant $\frac14$⁠, and everything else is a function of $x^4 = t$⁠.
   Integration by substitution, Section 7.3.1. Lesson: substitution

3. Split $\frac{1}{t(t - 1)}$

   $$\begin{gathered} \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1}, \\ 1 = A(t - 1) + Bt. \end{gathered}$$

   Put $t = 0$⁠: $1 = -A$⁠, $A = -1$⁠. Put $t = 1$⁠: $B = 1$⁠. So

   $$\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac1t.$$
   Two different linear factors $t - 0$ and $t - 1$⁠: row 1 of Table 7.2. Check at $t = 2$⁠: $\frac{1}{2} = 1 - \frac12$⁠. ✓
   Table 7.2, row 1. Basics: splitting fractions

4. Integrate and return to $x$

   $$\begin{aligned} \frac14\int\Big(\frac{1}{t - 1} - \frac1t\Big)dt &= \frac14\big(\log|t - 1| - \log|t|\big) + C \\ &= \frac14\log\left|\frac{t - 1}{t}\right| + C \\ &= \frac14\log\left|\frac{x^4 - 1}{x^4}\right| + C. \end{aligned}$$
   Each simple fraction integrates to a logarithm; the difference of logs is the log of the quotient; finally $t = x^4$⁠.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

5. Cross-check with the full decomposition

   Without the hint, factor $x(x^4 - 1) = x(x - 1)(x + 1)(x^2 + 1)$ and split:

   $$\begin{aligned} \frac{1}{x(x^4 - 1)} &= -\frac1x + \frac{1}{4(x - 1)} \\ &\quad + \frac{1}{4(x + 1)} + \frac{x}{2(x^2 + 1)}. \end{aligned}$$

   (The constants come from putting $x = 0, 1, -1$ in the cleared identity, and the last numerator from comparing coefficients.) Integrating,

   $$\begin{aligned} &-\log|x| + \tfrac14\log|x^2 - 1| + \tfrac14\log(x^2 + 1) \\ &= \tfrac14\log|x^4 - 1| - \tfrac14\log x^4, \end{aligned}$$

   which is the same answer.

   $\log|x - 1| + \log|x + 1| = \log|x^2 - 1|$⁠, then adding $\log(x^2 + 1)$ gives $\log|x^4 - 1|$⁠; and $\log|x| = \frac14\log x^4$⁠. Two independent routes agreeing is a strong check.
   Section 7.5. Basics: splitting fractions

6. Check: differentiate the answer

   Write the answer as $\frac14\log|x^4 - 1| - \log|x|$⁠. Then

   $$\begin{aligned} &\frac{d}{dx}\Big(\tfrac14\log|x^4 - 1| - \log|x|\Big) \\ &= \frac{x^3}{x^4 - 1} - \frac1x \\ &= \frac{x^4 - (x^4 - 1)}{x(x^4 - 1)} = \frac{1}{x(x^4 - 1)}. \ \checkmark \end{aligned}$$

   Valid for $x \ne 0, \pm 1$⁠.

   Chain rule: $\frac{d}{dx}\log|x^4 - 1| = \frac{4x^3}{x^4 - 1}$⁠, and $\frac14$ of it is $\frac{x^3}{x^4 - 1}$⁠. Over the common denominator the $x^4$ terms cancel and 1 remains.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

Every fact used, and where it comes from