Miscellaneous · Q6: $\int\frac{5x}{(x+1)(x^2+9)}\,dx$

Basics you need

Solution

1. Choose the method

   The numerator $5x$ has degree 1 and the denominator degree 3, so the fraction is proper. The factor $x^2 + 9$ has discriminant $0 - 36 < 0$⁠, so it has no real roots and cannot be factorised further.

   Neither a standard formula nor a single substitution fits the whole fraction (the numerator is not a multiple of the derivative of $(x+1)(x^2+9)$⁠). Partial fractions break it into pieces we can integrate, and row 5 of Table 7.2 is exactly the "linear times irreducible quadratic" case.
   Section 7.5, Table 7.2 row 5. Basics: splitting fractions

2. Write the partial fraction form

   $$\frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+9}.$$

   Multiply by $(x+1)(x^2+9)$⁠:

   $$5x = A(x^2 + 9) + (Bx + C)(x + 1). \quad (1)$$
   Over an irreducible quadratic the numerator must be allowed to be linear, $Bx + C$⁠, not just a constant. Equation (1) is a polynomial identity, true for every real $x$⁠.
   Table 7.2, row 5 (Basic 1).

3. Find $A$⁠, $B$ and $C$

   $x = -1$ (root of $x + 1$⁠): $-5 = A(1 + 9) = 10A$⁠, so $A = -\frac12$⁠.

   Expand the right side of (1):

   $$(A + B)x^2 + (B + C)x + (9A + C).$$

   Compare with $0x^2 + 5x + 0$⁠:

   $$\begin{aligned} x^2:&\quad A + B = 0 \ \Rightarrow\ B = \tfrac12, \\ \text{constant}:&\quad 9A + C = 0 \ \Rightarrow\ C = \tfrac92, \\ x:&\quad B + C = \tfrac12 + \tfrac92 = 5. \ \checkmark \end{aligned}$$

   So

   $$\frac{5x}{(x+1)(x^2+9)} = -\frac{1}{2(x+1)} + \frac{x + 9}{2(x^2+9)}.$$
   The root of $x + 1$ gives $A$ immediately; the quadratic has no real root to put in, so we compare coefficients for the rest. Three equations for two remaining unknowns: the third one is a free check.
   Equal polynomials have equal coefficients. Basics: splitting fractions

4. Integrate the linear piece

   $$\int -\frac{1}{2(x+1)}\,dx = -\frac12\log|x+1|.$$
   With $t = x + 1$⁠, $dt = dx$⁠, this is standard formula (xii). The absolute value stays, since $x + 1$ can be negative.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

5. Integrate the quadratic piece in two parts

   Split the numerator: $\dfrac{x + 9}{2(x^2+9)} = \dfrac12\cdot\dfrac{x}{x^2+9} + \dfrac92\cdot\dfrac{1}{x^2+9}$⁠.

   First part: $t = x^2 + 9$⁠, $dt = 2x\,dx$⁠:

   $$\frac12\int\frac{x\,dx}{x^2+9} = \frac12\cdot\frac12\int\frac{dt}{t} = \frac14\log(x^2+9).$$

   Second part: formula (3) of Section 7.4 with $a = 3$⁠:

   $$\frac92\int\frac{dx}{x^2+3^2} = \frac92\cdot\frac13\tan^{-1}\frac x3 = \frac32\tan^{-1}\frac x3.$$
   The $x$ in the numerator is half the derivative of $x^2 + 9$⁠, so that part is a logarithm; the constant part is a "$\frac{1}{x^2 + a^2}$" standard form. No absolute value is needed in $\log(x^2 + 9)$ because $x^2 + 9 > 0$⁠.
   Substitution, Section 7.3.1; formula (3), Section 7.4. Lesson: integrals of particular functions

6. Assemble the answer

   $$\begin{aligned} \int f(x)\,dx &= -\frac12\log|x+1| + \frac14\log(x^2+9) \\ &\quad + \frac32\tan^{-1}\frac x3 + C. \end{aligned}$$
   The three pieces from Steps 4 and 5 are added, with one constant $C$⁠. This is exactly the textbook's answer.
   Properties (III), (IV), Section 7.2.1.

7. Check: differentiate the answer

   Using $\frac{d}{dx}\tan^{-1}\frac x3 = \frac{1/3}{1 + x^2/9} = \frac{3}{x^2+9}$⁠:

   $$\begin{aligned} F' &= -\frac{1}{2(x+1)} + \frac{2x}{4(x^2+9)} + \frac32\cdot\frac{3}{x^2+9} \\ &= -\frac{1}{2(x+1)} + \frac{x + 9}{2(x^2+9)} \\ &= \frac{-(x^2+9) + (x+9)(x+1)}{2(x+1)(x^2+9)} \\ &= \frac{10x}{2(x+1)(x^2+9)} = \frac{5x}{(x+1)(x^2+9)}. \ \checkmark \end{aligned}$$
   $(x + 9)(x + 1) = x^2 + 10x + 9$⁠, and subtracting $x^2 + 9$ leaves $10x$⁠. The answer is valid on each of the intervals $x < -1$ and $x > -1$⁠.
   Chain rule; derivative of $\tan^{-1}$⁠, Chapter 5. Basics: integration undoes differentiation

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