Exercise 7.5 · Q11: $\int\frac{5x}{(x+1)(x^2-4)}\,dx$

Basics you need

Solution

1. Factor the denominator; the fraction is proper

   $$(x+1)(x^2-4) = (x+1)(x-2)(x+2).$$

   The numerator $5x$ has degree 1 and the denominator degree 3, so the fraction is proper. The roots $-1$⁠, $2$⁠, $-2$ are different.

   Partial fractions need the denominator fully factored into linear (or irreducible quadratic) factors. $x^2 - 4$ still factors, so it must be split.
   Section 7.5. Basics: splitting fractions

2. Write the form of the decomposition

   $$\begin{aligned} &\frac{5x}{(x+1)(x-2)(x+2)} \\ &\quad = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2} \qquad (1) \end{aligned}$$
   One constant over each different linear factor.
   Table 7.2, row 3. Lesson: partial fractions

3. Clear the denominators

   Multiply (1) by $(x+1)(x-2)(x+2)$⁠:

   $$\begin{aligned} 5x &= A(x-2)(x+2) \\ &\quad + B(x+1)(x+2) \\ &\quad + C(x+1)(x-2). \qquad (2) \end{aligned}$$
   (2) is a polynomial identity, so it holds for every real $x$⁠, including the three roots.
   Remark after Example 11, Section 7.5.

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

   $x = -1$⁠: $-5 = A(-3)(1) = -3A$⁠, so $A = \frac53$⁠.

   $x = 2$⁠: $10 = B(3)(4) = 12B$⁠, so $B = \frac{10}{12} = \frac56$⁠.

   $x = -2$⁠: $-10 = C(-1)(-4) = 4C$⁠, so $C = -\frac{10}{4} = -\frac52$⁠.

   At each root the other two terms contain a zero factor and vanish.

5. Confirm by comparing coefficients

   Expand: $(x-2)(x+2) = x^2 - 4$⁠, $(x+1)(x+2) = x^2 + 3x + 2$⁠, $(x+1)(x-2) = x^2 - x - 2$⁠. The right side of (2) is

   $$(A+B+C)x^2 + (3B - C)x + (-4A + 2B - 2C).$$ $$\begin{aligned} x^2:&\quad \tfrac53 + \tfrac56 - \tfrac52 = \tfrac{10 + 5 - 15}{6} = 0, \\ x:&\quad \tfrac52 + \tfrac52 = 5, \\ \text{const}:&\quad -\tfrac{20}{3} + \tfrac53 + 5 = 0. \end{aligned}$$
   These match $0x^2 + 5x + 0$⁠, so the constants are right.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the three simple fractions

   $$\begin{aligned} \int f(x)\,dx &= \frac53\int\frac{dx}{x+1} + \frac56\int\frac{dx}{x-2} \\ &\quad - \frac52\int\frac{dx}{x+2} \\ &= \frac53\log|x+1| - \frac52\log|x+2| \\ &\quad + \frac56\log|x-2| + C. \end{aligned}$$
   Integral of a sum, 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, a new constant, not the $C = -\frac52$ of the decomposition.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

7. Check: differentiate the answer

   $$F'(x) = \frac{5}{3(x+1)} - \frac{5}{2(x+2)} + \frac{5}{6(x-2)}.$$

   Over the common denominator $6(x+1)(x-2)(x+2)$ the numerator is

   $$\begin{aligned} &10(x^2-4) - 15(x^2-x-2) + 5(x^2+3x+2) \\ &= 0x^2 + 30x + 0 = 30x, \end{aligned}$$

   so $F'(x) = \frac{30x}{6(x+1)(x-2)(x+2)}$ $= \frac{5x}{(x+1)(x^2-4)}$⁠. ✓

   Each log term is multiplied by the two factors it lacks. The answer is valid on each interval avoiding $-1$⁠, $2$⁠, $-2$⁠.
   $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2, list item (xii). Basics: integration undoes differentiation

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