Miscellaneous · Q1: $\int\frac{dx}{x-x^3}$

Basics you need

Solution

1. Choose the method and factorise the denominator

   Take out the common factor $x$⁠, then use the difference of squares $1 - x^2 = (1-x)(1+x)$⁠:

   $$x - x^3 = x(1 - x^2) = x(1-x)(1+x).$$

   So the integrand is $\dfrac{1}{x(1-x)(1+x)}$⁠. It is defined for $x \ne 0, 1, -1$⁠.

   No standard formula fits $\frac{1}{x-x^3}$ directly, and the substitution $t = x - x^3$ does not work, because the numerator is not a multiple of the derivative of the denominator, which is $1 - 3x^2$⁠. But it is a rational function with a numerator of lower degree than the denominator (proper), and the denominator breaks into different linear factors. For that case the method of Section 7.5 always works.
   Proper rational functions and Table 7.2, Section 7.5. Basics: splitting fractions

2. Write the partial fraction form

   With three different linear factors, Table 7.2 row 3 gives

   $$\frac{1}{x(1-x)(1+x)} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}.$$

   Multiply both sides by $x(1-x)(1+x)$⁠:

   $$\begin{aligned} 1 &= A(1-x)(1+x) + Bx(1+x) \\ &\quad + Cx(1-x). \qquad (1) \end{aligned}$$
   The factor $1 - x$ is $-(x - 1)$⁠, so $\frac{B}{1-x}$ is the same kind of term as $\frac{-B}{x-1}$⁠; keeping the factor as $1 - x$ just saves a sign. Equation (1) holds for every $x$ other than $0, \pm1$⁠; both sides are polynomials, so it is a polynomial identity and holds for all real $x$⁠, including the roots we use next.
   Table 7.2, row 3, Section 7.5. Lesson: partial fractions

3. Find $A$⁠, $B$ and $C$ by putting in the roots

   $x = 0$⁠: $1 = A(1)(1)$⁠, so $A = 1$⁠.

   $x = 1$⁠: $1 = B(1)(2)$⁠, so $B = \frac12$⁠.

   $x = -1$⁠: $1 = C(-1)(2) = -2C$⁠, so $C = -\frac12$⁠.

   Hence

   $$\frac{1}{x-x^3} = \frac1x + \frac{1}{2(1-x)} - \frac{1}{2(1+x)}.$$
   Each root makes two of the three products in (1) vanish, leaving one equation in one unknown. Confirm by comparing coefficients: the right side of (1) is $(-A + B - C)x^2 + (B + C)x + A$⁠. With our values: $-1 + \frac12 + \frac12 = 0$⁠, $\frac12 - \frac12 = 0$ and $A = 1$⁠, which matches the left side $0x^2 + 0x + 1$⁠.
   A polynomial identity may be evaluated at any $x$⁠; equal polynomials have equal coefficients. Basics: splitting fractions

4. Integrate each piece

   For the middle term use $t = 1 - x$⁠, $dt = -dx$⁠; for the last, $t = 1 + x$⁠, $dt = dx$⁠:

   $$\int\frac{dx}{1-x} = -\log|1-x|, \qquad \int\frac{dx}{1+x} = \log|1+x|.$$

   So

   $$\begin{aligned} \int\frac{dx}{x-x^3} &= \int\frac{dx}{x} + \frac12\int\frac{dx}{1-x} \\ &\quad - \frac12\int\frac{dx}{1+x} \\ &= \log|x| - \frac12\log|1-x| \\ &\quad - \frac12\log|1+x| + C. \end{aligned}$$
   The integral of a sum is the sum of the integrals, and constants come outside (properties (III) and (IV)). The coefficient of $x$ in $1 - x$ is $-1$⁠, so that logarithm picks up a minus sign: this is the step where sign slips happen.
   Standard formula (xii), Section 7.2; substitution, Section 7.3.1. Basics: substitution

5. Combine the logarithms into the key's form

   Since $|1-x|\,|1+x| = |1-x^2|$⁠, the last two terms combine:

   $$-\tfrac12\big(\log|1-x| + \log|1+x|\big) = -\tfrac12\log|1-x^2|.$$

   Also $\log|x| = \frac12\log x^2$⁠. Therefore

   $$\begin{aligned} \int\frac{dx}{x-x^3} &= \frac12\log x^2 - \frac12\log|1-x^2| + C \\ &= \frac12\log\left|\frac{x^2}{1-x^2}\right| + C. \end{aligned}$$
   This is exactly the textbook's form, so the two answers are identical (the difference is 0, not just a constant). Each law of logarithms here is applied to positive numbers $|x|$⁠, $|1-x|$⁠, $|1+x|$⁠, which is why the absolute values are kept until the end.
   $\log p + \log q = \log pq$⁠, $k\log p = \log p^k$ for $p, q > 0$ (Basic 3).

6. Check: differentiate the answer

   Use $\frac{d}{dx}\log|u| = \frac{u'}{u}$ with $u = x$ and $u = 1 - x^2$⁠:

   $$\begin{aligned} &\frac{d}{dx}\Big(\log|x| - \tfrac12\log|1-x^2|\Big) \\ &= \frac1x - \frac12\cdot\frac{-2x}{1-x^2} = \frac1x + \frac{x}{1-x^2} \\ &= \frac{(1-x^2) + x^2}{x(1-x^2)} = \frac{1}{x-x^3}. \ \checkmark \end{aligned}$$
   The derivative returns the integrand, so the answer is right. The check is valid on each of the four intervals $x < -1$⁠, $-1 < x < 0$⁠, $0 < x < 1$⁠, $x > 1$⁠; on each one the constant $C$ may be different.
   Chain rule and $\frac{d}{dx}\log|u| = \frac{u'}{u}$⁠, Chapter 5. Basics: integration undoes differentiation

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