Exercise 7.5 · Q1: $\int\frac{x}{(x+1)(x+2)}\,dx$

Basics you need

Solution

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

   The numerator $x$ has degree 1. The denominator $(x+1)(x+2) = x^2 + 3x + 2$ has degree 2. Since $1 < 2$⁠, the fraction is proper. Its factors $x + 1 = x - (-1)$ and $x + 2 = x - (-2)$ are linear and different, so row 1 of Table 7.2 applies with $a = -1$⁠, $b = -2$⁠.

   The forms in Table 7.2 are only for proper fractions. If the numerator's degree were not smaller, a polynomial part would first have to be divided out (as in Q6 and Q12 of this exercise).
   Section 7.5 (proper and improper rational functions). Lesson: partial fractions

2. Write the form of the decomposition

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

   where $A$ and $B$ are real numbers still to be found.

   Each different linear factor of the denominator gets its own simple fraction with a constant on top. Adding $\frac{A}{x+1}$ and $\frac{B}{x+2}$ gives a fraction over $(x+1)(x+2)$ with a numerator of degree at most 1, which is exactly the kind of numerator we have, so suitable $A$⁠, $B$ exist.
   Table 7.2, row 1. Basics: splitting fractions

3. Clear the denominators

   Multiply both sides of (1) by $(x+1)(x+2)$⁠:

   $$x = A(x+2) + B(x+1). \qquad (2)$$
   This removes every fraction. Identity (1) holds for every $x \ne -1, -2$⁠, so the two polynomials in (2) agree at infinitely many points. Two polynomials that agree at infinitely many points are the same polynomial, so (2) holds for every real $x$⁠, including $-1$ and $-2$⁠.
   The book's Remark after Example 11 (Section 7.5): (1) is an identity. Basics: splitting fractions

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

   Put $x = -1$ in (2). The bracket $(x+1)$ becomes 0, so the $B$ term vanishes:

   $$-1 = A(-1+2) + 0 = A, \quad\text{so } A = -1.$$

   Put $x = -2$ in (2). Now the bracket $(x+2)$ is 0:

   $$-2 = 0 + B(-2+1) = -B, \quad\text{so } B = 2.$$
   Choosing $x$ to be a root of one factor makes that factor's term disappear and leaves a one-line equation for the other constant. This is allowed because (2) holds for every $x$ (Step 3).

5. Confirm by comparing coefficients

   Expand the right side of (2): $A(x+2) + B(x+1)$ $= (A+B)x + (2A+B)$⁠. Compare with $1\cdot x + 0$⁠:

   $$\begin{aligned} \text{coefficient of } x:&\quad A + B = 1, \\ \text{constant term}:&\quad 2A + B = 0. \end{aligned}$$

   With $A = -1$⁠, $B = 2$⁠: $-1 + 2 = 1$ and $-2 + 2 = 0$⁠. Both equations hold.

   This is the book's method (Example 11). It gives an independent check that the values from Step 4 are right. Solving the two equations directly gives the same answer: subtracting the first from the second, $A = -1$⁠, then $B = 2$⁠.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the two simple fractions

   $$\frac{x}{(x+1)(x+2)} = -\frac{1}{x+1} + \frac{2}{x+2},$$ $$\begin{aligned} \int\frac{x\,dx}{(x+1)(x+2)} &= -\int\frac{dx}{x+1} + 2\int\frac{dx}{x+2} \\ &= -\log|x+1| + 2\log|x+2| + C. \end{aligned}$$
   The integral of a sum is the sum of the integrals, and constants move outside (Properties (III) and (IV), Section 7.2.1). Each piece is $\int\frac{dx}{x+k}$⁠: substitute $t = x+k$⁠, $dt = dx$⁠, and use $\int\frac{dt}{t} = \log|t| + C$⁠.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

7. Write it as one logarithm (the textbook's form)

   Since $2\log|x+2| = \log|x+2|^2$ $= \log(x+2)^2$⁠:

   $$-\log|x+1| + 2\log|x+2| = \log\frac{(x+2)^2}{|x+1|}.$$
   $k\log p = \log p^k$ and $\log p - \log q = \log\frac pq$ for positive $p$⁠, $q$⁠. The absolute value on $x + 2$ can be dropped because $|x+2|^2 = (x+2)^2$⁠.
   Basics: logarithms and 1/x

8. Check: differentiate the answer

   Using $\frac{d}{dx}\log|x+k| = \frac{1}{x+k}$⁠:

   $$\begin{aligned} &\frac{d}{dx}\big(-\log|x+1| + 2\log|x+2|\big) \\ &= -\frac{1}{x+1} + \frac{2}{x+2} \\ &= \frac{-(x+2) + 2(x+1)}{(x+1)(x+2)} \\ &= \frac{x}{(x+1)(x+2)}. \ \checkmark \end{aligned}$$
   Over the common denominator the numerator is $-x - 2 + 2x + 2 = x$⁠, the original numerator. The answer is valid on every interval that avoids $-1$ and $-2$⁠.
   $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2, list item (xii). Basics: integration undoes differentiation

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