Exercise 7.5 · Q3: $\int\frac{3x-1}{(x-1)(x-2)(x-3)}\,dx$

Basics you need

Solution

1. The fraction is proper, with three different linear factors

   The numerator $3x - 1$ has degree 1; the denominator is a product of three linear factors, so it has degree 3. Since $1 < 3$⁠, the fraction is proper. The factors $x - 1$⁠, $x - 2$⁠, $x - 3$ are different ($a = 1$⁠, $b = 2$⁠, $c = 3$⁠).

   Row 3 of Table 7.2 covers a proper fraction over three different linear factors. The numerator $3x - 1$ is the case $p = 0$⁠, $q = 3$⁠, $r = -1$ of $px^2 + qx + r$⁠.
   Section 7.5. Lesson: partial fractions

2. Write the form of the decomposition

   $$\begin{aligned} &\frac{3x-1}{(x-1)(x-2)(x-3)} \\ &\quad = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \qquad (1) \end{aligned}$$
   One constant over each different linear factor. Three unknowns match the three coefficients (of $x^2$⁠, $x$⁠, and the constant) we will be able to compare.
   Table 7.2, row 3. Basics: splitting fractions

3. Clear the denominators

   Multiply (1) by $(x-1)(x-2)(x-3)$⁠. Each fraction loses its own factor and keeps the other two:

   $$\begin{aligned} 3x - 1 &= A(x-2)(x-3) \\ &\quad + B(x-1)(x-3) \\ &\quad + C(x-1)(x-2). \qquad (2) \end{aligned}$$
   (1) is an identity for $x \ne 1, 2, 3$⁠; both sides of (2) are polynomials agreeing at infinitely many points, so (2) holds for every real $x$⁠.
   Remark after Example 11, Section 7.5.

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

   $x = 1$⁠: only the $A$ term survives.

   $$3 - 1 = A(1-2)(1-3) \;\Rightarrow\; 2 = 2A \;\Rightarrow\; A = 1.$$

   $x = 2$⁠: only the $B$ term survives.

   $$6 - 1 = B(2-1)(2-3) \;\Rightarrow\; 5 = -B \;\Rightarrow\; B = -5.$$

   $x = 3$⁠: only the $C$ term survives.

   $$9 - 1 = C(3-1)(3-2) \;\Rightarrow\; 8 = 2C \;\Rightarrow\; C = 4.$$
   At $x = 1$ the factor $(x - 1)$ is zero, so the $B$ and $C$ terms, which both contain $(x-1)$⁠, vanish. The same happens at $x = 2$ and $x = 3$ for the other terms.

5. Confirm by comparing coefficients

   Expand the three products:

   $$\begin{aligned} (x-2)(x-3) &= x^2 - 5x + 6, \\ (x-1)(x-3) &= x^2 - 4x + 3, \\ (x-1)(x-2) &= x^2 - 3x + 2. \end{aligned}$$

   So the right side of (2) is $(A+B+C)x^2$ $- (5A+4B+3C)x$ $+ (6A+3B+2C)$⁠. Compare with $0x^2 + 3x - 1$⁠, using $A = 1$⁠, $B = -5$⁠, $C = 4$⁠:

   $$\begin{aligned} x^2:&\quad A+B+C = 1 - 5 + 4 = 0, \\ x:&\quad -(5A+4B+3C) = -(5 - 20 + 12) = 3, \\ \text{const}:&\quad 6A+3B+2C = 6 - 15 + 8 = -1. \end{aligned}$$
   All three coefficients match, so the constants are right. Comparing coefficients alone would mean solving three simultaneous equations; putting in the roots is faster, and this step is the check.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the three simple fractions

   $$\begin{aligned} &\frac{3x-1}{(x-1)(x-2)(x-3)} \\ &\quad = \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}, \end{aligned}$$ $$\begin{aligned} \int f(x)\,dx &= \int\frac{dx}{x-1} - 5\int\frac{dx}{x-2} \\ &\quad + 4\int\frac{dx}{x-3} \\ &= \log|x-1| - 5\log|x-2| \\ &\quad + 4\log|x-3| + C. \end{aligned}$$
   Split the integral over the sum and take the constants outside (Properties (III), (IV), Section 7.2.1). Each piece is $\int\frac{dx}{x+k} = \log|x+k|$⁠. The final $C$ is the arbitrary constant of integration, not the $C = 4$ of the decomposition.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

7. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\big(\log|x-1| - 5\log|x-2| + 4\log|x-3|\big) \\ &= \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}. \end{aligned}$$

   Over the common denominator $(x-1)(x-2)(x-3)$⁠, using the expansions of Step 5, the numerator is

   $$\begin{aligned} &(x^2-5x+6) - 5(x^2-4x+3) + 4(x^2-3x+2) \\ &= 0x^2 + 3x - 1 = 3x - 1. \ \checkmark \end{aligned}$$
   We get back the integrand $\frac{3x-1}{(x-1)(x-2)(x-3)}$⁠. The answer is valid on every interval that avoids $1$⁠, $2$ and $3$⁠.
   $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2, list item (xii). Basics: integration undoes differentiation

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