Exercise 7.5 · Q18: $\int\frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}\,dx$

Basics you need

Solution

1. Write everything in $y = x^2$

   With $y = x^2$⁠,

   $$\begin{aligned} \frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)} &= \frac{(y + 1)(y + 2)}{(y + 3)(y + 4)} \\ &= \frac{y^2 + 3y + 2}{y^2 + 7y + 12}. \end{aligned}$$
   Only even powers of $x$ appear, so treating $x^2$ as one quantity halves the degrees and makes the algebra easier. As after Example 14, $y$ is used only for the algebra; we still integrate with respect to $x$⁠.
   Example 14, Section 7.5. Lesson: partial fractions

2. Divide: the fraction is improper

   Numerator and denominator have the same degree, so subtract one copy of the denominator:

   $$\begin{aligned} \frac{y^2 + 3y + 2}{y^2 + 7y + 12} &= 1 + \frac{(y^2 + 3y + 2) - (y^2 + 7y + 12)}{y^2 + 7y + 12} \\ &= 1 - \frac{4y + 10}{(y + 3)(y + 4)}. \end{aligned}$$
   Partial fractions in the forms of Table 7.2 need a proper fraction. Long division of $y^2 + 3y + 2$ by $y^2 + 7y + 12$ gives quotient 1 and remainder $-4y - 10$⁠; the remainder has degree 1, less than 2, so the leftover fraction is proper.
   Improper rational functions, Section 7.5. Basics: splitting fractions

3. Split the proper part

   $$\begin{gathered} \frac{4y + 10}{(y + 3)(y + 4)} = \frac{A}{y + 3} + \frac{B}{y + 4}, \\ 4y + 10 = A(y + 4) + B(y + 3). \end{gathered}$$

   Put $y = -3$⁠: $-2 = A$⁠. Put $y = -4$⁠: $-6 = -B$⁠, so $B = 6$⁠. Coefficient check: $A + B = 4$ ✓ and $4A + 3B = -8 + 18 = 10$ ✓.

   The two linear factors $y + 3$ and $y + 4$ are different, so row 1 of Table 7.2 applies. Substituting the roots of the factors isolates each constant; the coefficient check confirms both.
   Table 7.2, row 1. Basics: splitting fractions

4. The integrand in its final form

   $$\begin{aligned} &1 - \Big(\frac{-2}{y + 3} + \frac{6}{y + 4}\Big) \\ &= 1 + \frac{2}{x^2 + 3} - \frac{6}{x^2 + 4}. \end{aligned}$$
   Distribute the minus sign over both pieces, then put $y = x^2$ back. Each piece is now a standard form: a constant, and two of the type $\frac{1}{x^2 + a^2}$⁠.

5. Integrate each piece

   Section 7.4, formula (3), with $a = \sqrt3$ and with $a = 2$⁠:

   $$\begin{aligned} \int\frac{2\,dx}{x^2 + 3} &= \frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}, \\ \int\frac{6\,dx}{x^2 + 4} &= \frac62\tan^{-1}\frac x2 = 3\tan^{-1}\frac x2. \end{aligned}$$

   So

   $$\begin{aligned} \int\frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}\,dx &= x + \frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} \\ &\quad - 3\tan^{-1}\frac x2 + C. \end{aligned}$$
   $x^2 + 3 = x^2 + (\sqrt3)^2$ and $x^2 + 4 = x^2 + 2^2$⁠, so formula (3) applies with these values of $a$⁠. The constant 1 integrates to $x$⁠. Constant multiples come outside the integral (Property (IV)).
   Section 7.4, formula (3); standard formula (i), Section 7.2. Lesson: integrals of particular functions

6. Check: differentiate the answer

   Chain rule: $\frac{d}{dx}\tan^{-1}\frac xa = \frac{1}{1 + x^2/a^2}\cdot\frac1a = \frac{a}{x^2 + a^2}$⁠. So

   $$\begin{aligned} \frac{d}{dx}\Big(\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}\Big) &= \frac{2}{\sqrt3}\cdot\frac{\sqrt3}{x^2 + 3} \\ &= \frac{2}{x^2 + 3}, \\ \frac{d}{dx}\Big(3\tan^{-1}\frac x2\Big) &= 3\cdot\frac{2}{x^2 + 4} \\ &= \frac{6}{x^2 + 4}. \end{aligned}$$

   The derivative of the answer is therefore $1 + \frac{2}{x^2 + 3} - \frac{6}{x^2 + 4}$⁠, and over the common denominator the numerator is

   $$\begin{aligned} &(x^2 + 3)(x^2 + 4) + 2(x^2 + 4) - 6(x^2 + 3) \\ &= x^4 + 7x^2 + 12 + 2x^2 + 8 - 6x^2 - 18 \\ &= x^4 + 3x^2 + 2 = (x^2 + 1)(x^2 + 2). \ \checkmark \end{aligned}$$
   This rebuilds the original fraction $\frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}$ for every real $x$⁠, which confirms both the division and the partial fractions.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

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