Exercise 7.5 · Q8: $\int\frac{x}{(x-1)^2(x+2)}\,dx$

Basics you need

Solution

1. Proper, with a repeated linear factor

   The numerator $x$ has degree 1; the denominator $(x-1)^2(x+2)$ has degree 3, so the fraction is proper. The factor $x - 1$ appears twice and $x + 2$ once.

   Row 4 of Table 7.2 covers exactly this: a proper fraction over $(x-a)^2(x-b)$⁠, here with $a = 1$⁠, $b = -2$⁠.
   Section 7.5. Lesson: partial fractions

2. Write the form of the decomposition

   $$\begin{aligned} &\frac{x}{(x-1)^2(x+2)} \\ &\quad = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2} \qquad (1) \end{aligned}$$
   A repeated factor $(x-1)^2$ needs both $\frac{A}{x-1}$ and $\frac{B}{(x-1)^2}$⁠. With only $\frac{B}{(x-1)^2} + \frac{C}{x+2}$ there would be two unknowns for three coefficients, and no solution in general.
   Table 7.2, row 4. Basics: splitting fractions

3. Clear the denominators

   Multiply (1) by $(x-1)^2(x+2)$⁠:

   $$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2. \qquad (2)$$
   $\frac{A}{x-1}$ loses one factor $x - 1$ and keeps $(x-1)(x+2)$⁠; $\frac{B}{(x-1)^2}$ keeps only $x + 2$⁠; $\frac{C}{x+2}$ keeps $(x-1)^2$⁠. As before, (2) holds for every real $x$ because it is a polynomial identity.
   Remark after Example 11, Section 7.5.

4. Find $B$ and $C$ by putting in the roots

   $x = 1$⁠: the $A$ and $C$ terms vanish.

   $$1 = B(1+2) = 3B, \quad\text{so } B = \tfrac13.$$

   $x = -2$⁠: the $A$ and $B$ terms vanish.

   $$-2 = C(-2-1)^2 = 9C, \quad\text{so } C = -\tfrac29.$$
   Both the $A$ and $C$ terms contain the factor $x - 1$⁠, so they vanish at $x = 1$⁠. No root isolates $A$⁠, so we need one more equation.

5. Find $A$ by comparing $x^2$ coefficients, then check

   Expand: $A(x-1)(x+2) = A(x^2 + x - 2)$⁠, $B(x+2)$⁠, $C(x-1)^2 = C(x^2 - 2x + 1)$⁠. The right side of (2) is

   $$(A + C)x^2 + (A + B - 2C)x + (-2A + 2B + C).$$

   The left side $x$ has no $x^2$ term, so $A + C = 0$ and $A = \frac29$⁠. Check the other two coefficients:

   $$\begin{aligned} x:&\quad \tfrac29 + \tfrac39 + \tfrac49 = 1, \\ \text{const}:&\quad -\tfrac49 + \tfrac69 - \tfrac29 = 0. \end{aligned}$$
   These match the left side $1x + 0$⁠, so all three constants are confirmed.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the three simple fractions

   $$\begin{aligned} \int f(x)\,dx &= \frac29\int\frac{dx}{x-1} + \frac13\int\frac{dx}{(x-1)^2} \\ &\quad - \frac29\int\frac{dx}{x+2}. \end{aligned}$$

   For the middle integral put $t = x - 1$⁠, $dt = dx$⁠:

   $$\int\frac{dx}{(x-1)^2} = \int t^{-2}\,dt = \frac{t^{-1}}{-1} = -\frac{1}{x-1}.$$

   So

   $$\begin{aligned} \int f(x)\,dx &= \frac29\log|x-1| - \frac{1}{3(x-1)} \\ &\quad - \frac29\log|x+2| + C \\ &= \frac29\log\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C. \end{aligned}$$
   $\frac{1}{(x-1)^2}$ is a power $t^{-2}$ with $n = -2 \ne -1$⁠, so the power rule applies and there is no logarithm. The two logs combine because $\log|p| - \log|q| = \log\left|\frac pq\right|$⁠. The final $C$ is the arbitrary constant of integration, not the $C = -\frac29$ of the decomposition.
   Standard formulae (i) and (xii), Section 7.2; substitution, Section 7.3.1. Basics: powers of x

7. Check: differentiate the answer

   Using $\frac{d}{dx}\big(-(x-1)^{-1}\big) = (x-1)^{-2}$⁠:

   $$F'(x) = \frac{2}{9(x-1)} + \frac{1}{3(x-1)^2} - \frac{2}{9(x+2)}.$$

   Over the common denominator $9(x-1)^2(x+2)$ the numerator is

   $$\begin{aligned} &2(x-1)(x+2) + 3(x+2) - 2(x-1)^2 \\ &= (2x^2 + 2x - 4) + (3x + 6) \\ &\quad - (2x^2 - 4x + 2) = 9x, \end{aligned}$$

   so $F'(x) = \frac{9x}{9(x-1)^2(x+2)}$ $= \frac{x}{(x-1)^2(x+2)}$⁠. ✓

   The derivative returns the integrand. The answer is valid on each interval avoiding $x = 1$ and $x = -2$⁠.
   Power rule and $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2. Basics: integration undoes differentiation

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