Exercise 7.5 · Q6: $\int\frac{1-x^2}{x(1-2x)}\,dx$

Basics you need

Solution

1. The fraction is improper

   The numerator $1 - x^2$ has degree 2. The denominator $x(1-2x) = x - 2x^2$ also has degree 2. Since the numerator's degree is not less than the denominator's, the fraction is improper.

   Table 7.2 only lists forms for proper fractions. Splitting an improper fraction straight into $\frac{A}{x} + \frac{B}{1-2x}$ would fail, because that sum is always proper and cannot equal an improper fraction.
   Section 7.5 (improper rational functions). Lesson: partial fractions

2. Divide: a constant plus a proper fraction

   Write both in descending powers: $1 - x^2 = -x^2 + 1$ and $x(1-2x) = -2x^2 + x$⁠. The leading terms give the quotient $\frac{-x^2}{-2x^2} = \frac12$⁠. Subtract $\frac12(-2x^2 + x) = -x^2 + \frac x2$⁠:

   $$(-x^2 + 1) - \left(-x^2 + \tfrac x2\right) = 1 - \tfrac x2 = \tfrac{2-x}{2}.$$

   So $1 - x^2 = \frac12\,x(1-2x) + \frac{2-x}{2}$⁠, and dividing by $x(1-2x)$⁠:

   $$\frac{1-x^2}{x(1-2x)} = \frac12 + \frac{2-x}{2x(1-2x)}. \qquad (1)$$
   The remainder $\frac{2-x}{2}$ has degree 1, less than 2, so the leftover fraction is proper. Check: $\frac12(x - 2x^2) + 1 - \frac x2 = 1 - x^2$⁠. ✓
   Long division of polynomials, Section 7.5. Basics: splitting fractions

3. Write the form for the proper part

   Leave the factor $\frac12$ aside and split

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

   Multiply by $x(1-2x)$⁠:

   $$2 - x = A(1-2x) + Bx. \qquad (3)$$
   $x$ and $1 - 2x$ are different linear factors, so row 1 of Table 7.2 applies. (3) holds for every $x \ne 0, \frac12$⁠, hence, being a polynomial identity, for every real $x$⁠.
   Table 7.2, row 1; Remark after Example 11, Section 7.5.

4. Find $A$ and $B$

   $x = 0$ (root of $x$⁠): $2 = A(1) + 0$⁠, so $A = 2$⁠.

   $x = \frac12$ (root of $1 - 2x$⁠): $2 - \frac12 = 0 + \frac B2$⁠, so $\frac32 = \frac B2$ and $B = 3$⁠.

   Confirm by coefficients: the right side of (3) is $(-2A + B)x + A$⁠:

   $$\begin{aligned} \text{coefficient of } x:&\quad -2A + B = -4 + 3 = -1, \\ \text{constant term}:&\quad A = 2. \end{aligned}$$
   These match the left side $-x + 2$⁠. Putting in $x = \frac12$ is allowed because (3) is a polynomial identity.
   Basics: splitting fractions

5. Assemble the integrand

   From (2), $\frac{2-x}{x(1-2x)} = \frac2x + \frac{3}{1-2x}$⁠. Put this into (1) and multiply by $\frac12$⁠:

   $$\frac{1-x^2}{x(1-2x)} = \frac12 + \frac1x + \frac{3}{2(1-2x)}.$$
   Half of $\frac2x$ is $\frac1x$⁠, and half of $\frac{3}{1-2x}$ is $\frac{3}{2(1-2x)}$⁠.

6. Integrate term by term

   For the last term substitute $t = 1 - 2x$⁠, $dt = -2\,dx$⁠, so $dx = -\frac{dt}{2}$⁠:

   $$\int\frac{dx}{1-2x} = -\frac12\int\frac{dt}{t} = -\frac12\log|1-2x| + C.$$

   Hence

   $$\begin{aligned} \int f(x)\,dx &= \int\tfrac12\,dx + \int\frac{dx}{x} \\ &\quad + \frac32\int\frac{dx}{1-2x} \\ &= \frac x2 + \log|x| \\ &\quad + \frac32\left(-\frac12\log|1-2x|\right) + C \\ &= \frac x2 + \log|x| - \frac34\log|1-2x| + C. \end{aligned}$$
   The coefficient of $x$ in $1 - 2x$ is $-2$⁠, so the logarithm picks up a factor $-\frac12$⁠. Forgetting this factor (or its minus sign) is the usual error here.
   Substitution, Section 7.3.1; standard formulae (i) and (xii), Section 7.2. Basics: substitution

7. Check: differentiate the answer

   By the chain rule, $\frac{d}{dx}\log|1-2x| = \frac{-2}{1-2x}$⁠. So

   $$\begin{aligned} &\frac{d}{dx}\Big(\frac x2 + \log|x| - \frac34\log|1-2x|\Big) \\ &= \frac12 + \frac1x + \frac{3}{2(1-2x)} \\ &= \frac{x(1-2x) + 2(1-2x) + 3x}{2x(1-2x)} \\ &= \frac{2 - 2x^2}{2x(1-2x)} = \frac{1-x^2}{x(1-2x)}. \ \checkmark \end{aligned}$$
   The numerator is $x - 2x^2 + 2 - 4x + 3x = 2 - 2x^2$⁠. Cancelling the 2 gives the integrand. Valid on each interval avoiding $0$ and $\frac12$⁠.
   Chain rule, Chapter 5. Basics: integration undoes differentiation

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