Exercise 7.5 · Q9: $\int\frac{3x+5}{x^3-x^2-x+1}\,dx$

Basics you need

Solution

1. Factorise the denominator by grouping

   $$\begin{aligned} x^3 - x^2 - x + 1 &= x^2(x-1) - 1(x-1) \\ &= (x-1)(x^2-1) \\ &= (x-1)(x-1)(x+1) \\ &= (x-1)^2(x+1). \end{aligned}$$

   The numerator $3x + 5$ has degree 1 and the denominator degree 3, so the fraction is proper.

   The first two terms share $x^2$⁠; the last two share $-1$⁠; both leave the bracket $x - 1$⁠. Then $x^2 - 1 = (x-1)(x+1)$ is a difference of squares. Check: $x = 1$ gives $1 - 1 - 1 + 1 = 0$ and $x = -1$ gives $-1 - 1 + 1 + 1 = 0$⁠, so both are roots.
   Algebra; Section 7.5. Basics: splitting fractions

2. Write the form of the decomposition

   $$\begin{aligned} &\frac{3x+5}{(x-1)^2(x+1)} \\ &\quad = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} \qquad (1) \end{aligned}$$
   The repeated factor $(x-1)^2$ needs a fraction for each power, $x - 1$ and $(x-1)^2$⁠; the single factor $x + 1$ gets one.
   Table 7.2, row 4 ($a = 1$⁠, $b = -1$⁠). Lesson: partial fractions

3. Clear the denominators

   Multiply (1) by $(x-1)^2(x+1)$⁠:

   $$\begin{aligned} 3x + 5 &= A(x-1)(x+1) + B(x+1) \\ &\quad + C(x-1)^2. \qquad (2) \end{aligned}$$
   Each fraction keeps the factors it does not have. (2) is a polynomial identity, valid for every real $x$⁠.
   Remark after Example 11, Section 7.5.

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

   $x = 1$⁠: $3 + 5 = B(2)$⁠, so $8 = 2B$ and $B = 4$⁠.

   $x = -1$⁠: $-3 + 5 = C(-2)^2$⁠, so $2 = 4C$ and $C = \frac12$⁠.

   At $x = 1$ the $A$ and $C$ terms contain $x - 1 = 0$⁠; at $x = -1$ the $A$ and $B$ terms contain $x + 1 = 0$⁠.

5. Find $A$⁠, then check every coefficient

   Expand: $A(x^2 - 1) + B(x+1) + C(x^2 - 2x + 1)$⁠. The right side of (2) is

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

   No $x^2$ on the left, so $A + C = 0$ and $A = -\frac12$⁠. Check:

   $$\begin{aligned} x:&\quad B - 2C = 4 - 1 = 3, \\ \text{const}:&\quad -A + B + C = \tfrac12 + 4 + \tfrac12 = 5. \end{aligned}$$
   These match $3x + 5$⁠, so $A = -\frac12$⁠, $B = 4$⁠, $C = \frac12$ are confirmed.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the three simple fractions

   $$\begin{aligned} \int f(x)\,dx &= -\frac12\int\frac{dx}{x-1} + 4\int\frac{dx}{(x-1)^2} \\ &\quad + \frac12\int\frac{dx}{x+1} \\ &= -\frac12\log|x-1| - \frac{4}{x-1} \\ &\quad + \frac12\log|x+1| + C \\ &= \frac12\log\left|\frac{x+1}{x-1}\right| - \frac{4}{x-1} + C. \end{aligned}$$
   With $t = x - 1$⁠, $\int t^{-2}\,dt = -t^{-1}$⁠, so $4\int\frac{dx}{(x-1)^2} = -\frac{4}{x-1}$⁠. The two logs combine as $\frac12(\log|x+1| - \log|x-1|)$⁠. The final $C$ is the arbitrary constant of integration, not the $C = \frac12$ of the decomposition.
   Standard formulae (i) and (xii), Section 7.2. Basics: logarithms and 1/x

7. Check: differentiate the answer

   $$F'(x) = -\frac{1}{2(x-1)} + \frac{4}{(x-1)^2} + \frac{1}{2(x+1)}.$$

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

   $$\begin{aligned} &-(x-1)(x+1) + 8(x+1) + (x-1)^2 \\ &= (-x^2 + 1) + (8x + 8) + (x^2 - 2x + 1) \\ &= 6x + 10, \end{aligned}$$

   so $F'(x) = \frac{6x+10}{2(x-1)^2(x+1)}$ $= \frac{3x+5}{(x-1)^2(x+1)}$⁠. ✓

   $\frac{d}{dx}\left(-\frac{4}{x-1}\right) = \frac{4}{(x-1)^2}$⁠. We get back the integrand, since $(x-1)^2(x+1) = x^3 - x^2 - x + 1$⁠. Valid on each interval avoiding $\pm 1$⁠.
   Power rule and $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2. Basics: integration undoes differentiation

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