Exercise 7.5 · Q10: $\int\frac{2x-3}{(x^2-1)(2x+3)}\,dx$

Basics you need

Solution

1. Factor the denominator; the fraction is proper

   $$(x^2-1)(2x+3) = (x-1)(x+1)(2x+3).$$

   The numerator $2x - 3$ has degree 1 and the denominator degree 3, so the fraction is proper. The three linear factors have different roots $1$⁠, $-1$⁠, $-\frac32$⁠.

   $x^2 - 1$ is a difference of squares. Different roots mean three different linear factors, which is row 3 of Table 7.2.
   Section 7.5. Basics: splitting fractions

2. Write the form of the decomposition

   $$\begin{aligned} &\frac{2x-3}{(x-1)(x+1)(2x+3)} \\ &\quad = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3} \qquad (1) \end{aligned}$$
   $\frac{C}{2x+3} = \frac{C/2}{x + \frac32}$⁠, so this is row 3 with $c = -\frac32$⁠. Keeping $2x + 3$ avoids fractions inside the factor.
   Table 7.2, row 3. Lesson: partial fractions

3. Clear the denominators

   Multiply (1) by $(x-1)(x+1)(2x+3)$⁠:

   $$\begin{aligned} 2x - 3 &= A(x+1)(2x+3) \\ &\quad + B(x-1)(2x+3) \\ &\quad + C(x-1)(x+1). \qquad (2) \end{aligned}$$
   A polynomial identity: 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

   $x = 1$⁠: $2 - 3 = A(2)(5)$⁠, so $-1 = 10A$ and $A = -\frac{1}{10}$⁠.

   $x = -1$⁠: $-2 - 3 = B(-2)(1)$⁠, so $-5 = -2B$ and $B = \frac52$⁠.

   $x = -\frac32$⁠: the root of $2x + 3$⁠.

   $$\begin{aligned} 2\left(-\tfrac32\right) - 3 &= C\left(-\tfrac52\right)\left(-\tfrac12\right), \\ -6 &= \tfrac54\,C, \quad\text{so } C = -\tfrac{24}{5}. \end{aligned}$$
   At each root two of the three terms vanish. At $x = -\frac32$⁠: $x - 1 = -\frac52$ and $x + 1 = -\frac12$⁠, whose product is $\frac54$⁠.

5. Confirm by comparing coefficients

   Expand: $(x+1)(2x+3) = 2x^2 + 5x + 3$⁠, $(x-1)(2x+3) = 2x^2 + x - 3$⁠, $(x-1)(x+1) = x^2 - 1$⁠. The right side of (2) is

   $$(2A+2B+C)x^2 + (5A+B)x + (3A-3B-C).$$

   With $A = -\frac1{10}$⁠, $B = \frac52$⁠, $C = -\frac{24}{5}$⁠:

   $$\begin{aligned} x^2:&\quad -\tfrac15 + 5 - \tfrac{24}{5} = 0, \\ x:&\quad -\tfrac12 + \tfrac52 = 2, \\ \text{const}:&\quad -\tfrac{3}{10} - \tfrac{15}{2} + \tfrac{24}{5} = -3. \end{aligned}$$
   These match $0x^2 + 2x - 3$⁠. (For the constant: $\frac{-3 - 75 + 48}{10} = -3$⁠.)
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the three simple fractions

   For the last one put $t = 2x + 3$⁠, $dt = 2\,dx$⁠:

   $$\int\frac{dx}{2x+3} = \frac12\int\frac{dt}{t} = \frac12\log|2x+3| + C.$$

   Hence

   $$\begin{aligned} \int f(x)\,dx &= -\frac1{10}\log|x-1| + \frac52\log|x+1| \\ &\quad - \frac{24}{5}\cdot\frac12\log|2x+3| + C \\ &= \frac52\log|x+1| - \frac1{10}\log|x-1| \\ &\quad - \frac{12}{5}\log|2x+3| + C. \end{aligned}$$
   Integral of a sum, constants outside (Properties (III), (IV), Section 7.2.1). The factor $\frac12$ from $dt = 2\,dx$ turns $-\frac{24}{5}$ into $-\frac{12}{5}$⁠. The final $C$ is the arbitrary constant of integration, not the $C = -\frac{24}{5}$ of the decomposition.
   Substitution, Section 7.3.1; standard formula (xii), Section 7.2. Basics: substitution

7. Check: differentiate the answer

   $$F'(x) = \frac{5}{2(x+1)} - \frac{1}{10(x-1)} - \frac{24}{5(2x+3)}.$$

   Over the common denominator $10(x-1)(x+1)(2x+3)$ the numerator is

   $$\begin{aligned} &25(x-1)(2x+3) - (x+1)(2x+3) \\ &\quad - 48(x-1)(x+1) \\ &= 25(2x^2+x-3) - (2x^2+5x+3) \\ &\quad - 48(x^2-1) \\ &= 0x^2 + 20x - 30 = 10(2x-3). \end{aligned}$$

   So $F'(x) = \frac{10(2x-3)}{10(x-1)(x+1)(2x+3)}$ $= f(x)$⁠. ✓

   $\frac{d}{dx}\log|2x+3| = \frac{2}{2x+3}$ (chain rule), so $-\frac{12}{5}\cdot\frac{2}{2x+3} = -\frac{24}{5(2x+3)}$⁠. The answer is valid on each interval avoiding $1$⁠, $-1$⁠, $-\frac32$⁠.
   Chain rule, Chapter 5; $\frac{d}{dx}\log|x| = \frac1x$⁠. Basics: integration undoes differentiation

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