Example 11: $\int\frac{dx}{(x + 1)(x + 2)}$ by partial fractions

Basics you need

Solution

1. The integrand is proper, with distinct linear factors

   The numerator 1 has degree 0; the denominator $(x + 1)(x + 2) = x^2 + 3x + 2$ has degree 2. So the fraction is proper. The factors $x + 1$ and $x + 2$ are linear and different ($a = -1$⁠, $b = -2$ in Table 7.2).

   Partial fractions in the form of Table 7.2 need a proper fraction; otherwise a polynomial part would be missing from the decomposition. Distinct linear factors mean row 1 of the table applies.
   Section 7.5. Lesson: partial fractions

2. Write the form of the decomposition

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

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

   Each linear factor of the denominator gets its own simple fraction with a constant on top. (The book cites this row as "Table 7.2 (i)"; in the table it is printed as row 1.)
   Table 7.2, row 1. Basics: splitting fractions

3. Clear the denominators

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

   $$1 = A(x + 2) + B(x + 1).$$
   This removes every fraction. As the book's Remark says, (1) is an identity: it must hold for every permissible $x$⁠, so this polynomial equation holds for all $x$ (both sides are polynomials that agree at infinitely many points).

4. Find $A$ and $B$ by comparing coefficients

   The right side is $(A + B)x + (2A + B)$⁠. Compare with $0\cdot x + 1$⁠:

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

   Subtracting the first from the second: $A = 1$⁠; then $B = -1$⁠.

   Two equal polynomials have equal coefficients. The left side, 1, has no $x$ term, so the $x$-coefficient on the right must be 0.
   Basics: splitting fractions

5. A quicker cross-check: substitute the roots

   Put $x = -1$ in $1 = A(x + 2) + B(x + 1)$⁠: $1 = A(1) + 0$⁠, so $A = 1$⁠. Put $x = -2$⁠: $1 = 0 + B(-1)$⁠, so $B = -1$⁠. Same values.

   Each root makes one bracket vanish and isolates the other constant. The polynomial identity of Step 3 holds for all $x$⁠, including $-1$ and $-2$⁠, even though the original fraction is not defined there.

6. Integrate the two simple fractions

   $$\frac{1}{(x + 1)(x + 2)} = \frac{1}{x + 1} - \frac{1}{x + 2},$$ $$\begin{aligned} \int\frac{dx}{(x + 1)(x + 2)} &= \int\frac{dx}{x + 1} - \int\frac{dx}{x + 2} \\ &= \log|x + 1| - \log|x + 2| + C \\ &= \log\left|\frac{x + 1}{x + 2}\right| + C. \end{aligned}$$
   $\int\frac{dx}{x + k} = \log|x + k|$⁠: put $t = x + k$⁠, $dt = dx$⁠, and use $\int\frac{dt}{t} = \log|t|$⁠. The difference of two logs is the log of the quotient, and $\frac{|x + 1|}{|x + 2|} = \left|\frac{x + 1}{x + 2}\right|$⁠.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

7. Check: differentiate the answer

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

   Valid for every $x \ne -1, -2$⁠.

   Over the common denominator the numerators give $(x + 2) - (x + 1) = 1$⁠, the original numerator.

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