Exercise 7.5 · Q5: $\int\frac{2x}{x^2+3x+2}\,dx$

Basics you need

Solution

1. Factor the denominator; the fraction is proper

   $$x^2 + 3x + 2 = (x+1)(x+2),$$

   since $(x+1)(x+2) = x^2 + 2x + x + 2$ $= x^2 + 3x + 2$⁠. The numerator $2x$ has degree 1, less than 2, so the fraction is proper, and the two linear factors are different.

   Partial fractions work on the factors of the denominator, so the denominator must first be factorised. Expanding the product back confirms the factorisation.
   Section 7.5. Basics: splitting fractions

2. Write the form of the decomposition

   $$\frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \qquad (1)$$
   One unknown constant over each different linear factor.
   Table 7.2, row 1. Lesson: partial fractions

3. Clear the denominators

   Multiply (1) by $(x+1)(x+2)$⁠:

   $$2x = A(x+2) + B(x+1). \qquad (2)$$
   (1) holds for every $x \ne -1, -2$⁠, so the polynomials in (2) agree at infinitely many points; hence (2) holds for every real $x$⁠.
   Remark after Example 11, Section 7.5.

4. Find $A$ and $B$

   $x = -1$⁠: $2(-1) = A(1) + 0$⁠, so $A = -2$⁠.

   $x = -2$⁠: $2(-2) = 0 + B(-1)$⁠, so $-4 = -B$ and $B = 4$⁠.

   Confirm by comparing coefficients: the right side of (2) is $(A+B)x + (2A+B)$⁠, and the left side is $2x + 0$⁠:

   $$\begin{aligned} \text{coefficient of } x:&\quad A + B = -2 + 4 = 2, \\ \text{constant term}:&\quad 2A + B = -4 + 4 = 0. \end{aligned}$$
   Putting in a root removes one term; the coefficient comparison is an independent check of both values.
   Basics: splitting fractions

5. Integrate the two simple fractions

   $$\frac{2x}{x^2+3x+2} = -\frac{2}{x+1} + \frac{4}{x+2},$$ $$\begin{aligned} \int\frac{2x\,dx}{x^2+3x+2} &= -2\int\frac{dx}{x+1} + 4\int\frac{dx}{x+2} \\ &= 4\log|x+2| - 2\log|x+1| + C. \end{aligned}$$
   Integral of a sum, constants outside (Properties (III), (IV), Section 7.2.1); each piece is $\int\frac{dx}{x+k} = \log|x+k|$⁠.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

6. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\big(4\log|x+2| - 2\log|x+1|\big) \\ &= \frac{4}{x+2} - \frac{2}{x+1} \\ &= \frac{4(x+1) - 2(x+2)}{(x+1)(x+2)} \\ &= \frac{2x}{x^2+3x+2}. \ \checkmark \end{aligned}$$
   The numerator $4x + 4 - 2x - 4 = 2x$ is the original numerator. The answer is valid on every interval avoiding $-1$ and $-2$⁠.
   $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2, list item (xii). Basics: integration undoes differentiation

7. A link with Q1

   The integrand here is exactly twice the integrand of Q1, $\frac{x}{(x+1)(x+2)}$⁠, whose integral was $2\log|x+2| - \log|x+1| + C$⁠. Twice that is $4\log|x+2| - 2\log|x+1|$⁠, matching our answer.

   A constant factor comes straight through an integral (Property (IV), Section 7.2.1), so the two answers must be in the ratio 2 : 1. This is a free consistency check.
   Property (IV), Section 7.2.1.

Every fact used, and where it comes from