Exercise 7.5 · Q14: $\int\frac{3x - 1}{(x + 2)^2}\,dx$

Basics you need

Solution

1. The fraction is proper with a repeated linear factor

   The numerator $3x - 1$ has degree 1; the denominator $(x + 2)^2 = x^2 + 4x + 4$ has degree 2. So the fraction is proper. Its denominator is the single linear factor $x + 2$ taken twice.

   A proper fraction needs no long division. A squared linear factor is row 2 of Table 7.2 (with $a = -2$⁠), not row 1, because there is only one distinct factor.
   Section 7.5. Lesson: partial fractions

2. Write the form of the decomposition

   $$\frac{3x - 1}{(x + 2)^2} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2} \qquad (1)$$
   A repeated factor $(x + 2)^2$ needs a fraction for each power that appears: $x + 2$ and $(x + 2)^2$⁠. Writing only $\frac{A}{(x + 2)^2}$ would force the numerator to be a constant, which $3x - 1$ is not.
   Table 7.2, row 2. Basics: splitting fractions

3. Clear the denominators and find $A$⁠, $B$

   Multiply (1) by $(x + 2)^2$⁠:

   $$3x - 1 = A(x + 2) + B = Ax + (2A + B).$$

   Compare coefficients:

   $$\begin{aligned} x:&\quad A = 3, \\ \text{constant}:&\quad 2A + B = -1. \end{aligned}$$

   So $B = -1 - 2(3) = -7$⁠.

   Cross-check: put $x = -2$ in the cleared identity: $3(-2) - 1 = 0 + B$⁠, so $B = -7$⁠. ✓

   (1) holds for every $x \ne -2$⁠, so the two polynomials in the cleared equation agree at infinitely many points and are therefore the same polynomial: the cleared equation holds for every real $x$ (including $x = -2$⁠), and equal polynomials have equal coefficients. Equivalently, $3x - 1 = 3(x + 2) - 7$⁠: rewriting the numerator in terms of $x + 2$ gives the same split at once.
   Basics: splitting fractions

4. Integrate the two pieces

   $$\frac{3x - 1}{(x + 2)^2} = \frac{3}{x + 2} - \frac{7}{(x + 2)^2}.$$

   Substitute $t = x + 2$⁠, $dt = dx$⁠:

   $$\begin{aligned} \int\frac{3\,dx}{x + 2} &= 3\log|x + 2| + C_1, \\ \int\frac{7\,dx}{(x + 2)^2} &= 7\int t^{-2}\,dt \\ &= -\frac{7}{x + 2} + C_2. \end{aligned}$$

   Subtracting the second from the first:

   $$\int\frac{3x - 1}{(x + 2)^2}\,dx = 3\log|x + 2| + \frac{7}{x + 2} + C.$$
   $\int t^{-2}\,dt = \frac{t^{-1}}{-1} = -\frac1t$ (power rule with $n = -2$⁠, allowed because only $n = -1$ is excluded). Subtracting $-\frac{7}{x + 2}$ gives $+\frac{7}{x + 2}$⁠. The squared piece gives no logarithm.
   Standard formulae (i) and (xii), Section 7.2; substitution, Section 7.3.1. Basics: powers of x

5. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\Big(3\log|x + 2| + \frac{7}{x + 2}\Big) \\ &= \frac{3}{x + 2} - \frac{7}{(x + 2)^2} \\ &= \frac{3(x + 2) - 7}{(x + 2)^2} = \frac{3x - 1}{(x + 2)^2}. \ \checkmark \end{aligned}$$

   Valid for every $x \ne -2$⁠.

   $\frac{d}{dx}(x + 2)^{-1} = -(x + 2)^{-2}$ by the chain rule (inner derivative 1). Over the common denominator, $3(x + 2) - 7 = 3x - 1$⁠, the original numerator.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

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