Example 13: $\int\frac{3x-2}{(x+1)^2(x+3)}\,dx$

Basics you need

Solution

1. Choose the form of the partial fractions

   The top $3x - 2$ has degree 1 and the bottom $(x+1)^2(x+3)$ has degree 3, so the fraction is proper and no division is needed. The bottom has the repeated linear factor $(x+1)^2$ and the single factor $(x+3)$⁠. By Table 7.2, row 4,

   $$\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}.$$
   With only $\frac{A}{(x+1)^2} + \frac{C}{x+3}$ the combined top would be $A(x+3) + C(x+1)^2$⁠, and there are not enough free constants to match every top of degree up to 2. Each power of a repeated factor gets its own fraction: three unknowns for a cubic bottom.
   Row 4 of Table 7.2 is written for $(x-a)^2(x-b)$⁠; here $a = -1$ and $b = -3$⁠. Lesson: partial fractions

2. Clear the denominators

   Multiply both sides by $(x+1)^2(x+3)$⁠:

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

   Expanding the brackets,

   $$\begin{aligned} 3x - 2 &= A(x^2 + 4x + 3) + B(x+3) \\ &\quad + C(x^2 + 2x + 1). \end{aligned}$$
   The term $\frac{A}{x+1}$ needs the missing factors $(x+1)(x+3)$⁠; $\frac{B}{(x+1)^2}$ needs only $(x+3)$⁠; $\frac{C}{x+3}$ needs $(x+1)^2$⁠. This identity must hold for every $x$⁠.
   $(x+1)(x+3) = x^2 + 4x + 3$ and $(x+1)^2 = x^2 + 2x + 1$⁠. Basics: splitting fractions

3. Find $A$⁠, $B$ and $C$

   Comparing the coefficients of $x^2$⁠, $x$ and the constant terms (the left side has $0x^2$⁠):

   $$\begin{aligned} x^2:&\quad A + C = 0, \\ x:&\quad 4A + B + 2C = 3, \\ 1:&\quad 3A + 3B + C = -2. \end{aligned}$$

   Faster: put $x = -1$ in the identity of Step 2. Only the $B$ term survives: $-3 - 2 = B\cdot 2$⁠, so $B = -\frac52$⁠. Put $x = -3$⁠: only the $C$ term survives: $-9 - 2 = C\cdot 4$⁠, so $C = -\frac{11}{4}$⁠. Then $A + C = 0$ gives $A = \frac{11}{4}$⁠.

   The remaining equations agree: $4\cdot\frac{11}{4} - \frac52 - \frac{11}{2} = 11 - 8 = 3$⁠, and $\frac{33}{4} - \frac{15}{2} - \frac{11}{4} = \frac{33 - 30 - 11}{4} = -2$⁠. So

   $$A = \frac{11}{4}, \quad B = -\frac{5}{2}, \quad C = -\frac{11}{4},$$ $$\begin{aligned} &\frac{3x-2}{(x+1)^2(x+3)} \\ &= \frac{11}{4(x+1)} - \frac{5}{2(x+1)^2} - \frac{11}{4(x+3)}. \end{aligned}$$
   The two roots give $B$ and $C$ at once; the repeated root cannot give $A$ by itself (every $A$ term vanishes at both roots), so we take $A$ from the $x^2$ coefficient. Checking the other two equations confirms all three values together, which is the book's result.
   Substituting $x = -1$ or $x = -3$ is valid: both sides of the cleared identity are polynomials that agree for every other $x$⁠, hence they are the same polynomial and agree at $-1$ and $-3$ too. Lesson: partial fractions

4. Integrate each piece

   $$\begin{aligned} \int \frac{3x-2}{(x+1)^2(x+3)}\,dx &= \frac{11}{4}\int\frac{dx}{x+1} - \frac52\int\frac{dx}{(x+1)^2} \\ &\quad - \frac{11}{4}\int\frac{dx}{x+3}. \end{aligned}$$

   The first and last are logarithms: $\log|x+1|$ and $\log|x+3|$⁠. For the middle one put $t = x + 1$⁠, $dt = dx$⁠:

   $$\int \frac{dx}{(x+1)^2} = \int t^{-2}\,dt = \frac{t^{-1}}{-1} = -\frac{1}{x+1}.$$

   So, with $-\frac52 \cdot \left(-\frac{1}{x+1}\right) = +\frac{5}{2(x+1)}$⁠,

   $$\begin{aligned} \int \frac{3x-2}{(x+1)^2(x+3)}\,dx &= \frac{11}{4}\log|x+1| + \frac{5}{2(x+1)} \\ &\quad - \frac{11}{4}\log|x+3| + C. \end{aligned}$$
   The squared piece is not a logarithm: $\frac{1}{(x+1)^2}$ is the power $(x+1)^{-2}$⁠, and the power rule applies because $n = -2 \ne -1$⁠. The two minus signs multiply to a plus, a common place for a slip.
   Formula (i), power rule, and formula (xii), with the linear substitution $t = x + a$⁠. Basics: powers of x

5. Combine the logarithms

   Using $\log|u| - \log|v| = \log\left|\frac uv\right|$⁠:

   $$\begin{aligned} &\int \frac{3x-2}{(x+1)^2(x+3)}\,dx \\ &= \frac{11}{4}\log\left|\frac{x+1}{x+3}\right| + \frac{5}{2(x+1)} + C. \end{aligned}$$
   Both log terms carry the same coefficient $\frac{11}{4}$⁠, so they combine into one log of a quotient. This is the form the book gives.
   Quotient law of logarithms. Basics: logarithms and 1/x

6. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left[\frac{11}{4}\log|x+1| + \frac{5}{2(x+1)} - \frac{11}{4}\log|x+3|\right] \\ &= \frac{11}{4(x+1)} - \frac{5}{2(x+1)^2} - \frac{11}{4(x+3)}. \end{aligned}$$

   Over the common denominator $4(x+1)^2(x+3)$ the top is

   $$\begin{aligned} &11(x+1)(x+3) - 10(x+3) - 11(x+1)^2 \\ &= 11x^2 + 44x + 33 - 10x - 30 \\ &\quad - 11x^2 - 22x - 11 \\ &= 12x - 8 = 4(3x - 2). \end{aligned}$$

   Dividing by $4(x+1)^2(x+3)$ gives $\frac{3x-2}{(x+1)^2(x+3)}$⁠, the integrand.

   The $x^2$ terms cancel and the factor 4 cancels, leaving exactly the original integrand, so the antiderivative is right.
   $\frac{d}{dx}\frac{1}{x+1} = -\frac{1}{(x+1)^2}$ and $\frac{d}{dx}\log|u| = \frac1u$⁠. Basics: integration undoes differentiation

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