Miscellaneous · Q14: $\int\frac{dx}{(x^2+1)(x^2+4)}$

Basics you need

Solution

1. Choose the method

   The integrand is a proper rational function (numerator degree 0, denominator degree 4), and its denominator is already factorised into $x^2+1$ and $x^2+4$⁠. Neither factor has a real root. Table 7.2 has no row for two quadratic factors, but following its row 5 (a quadratic factor that cannot be factorised gets a numerator of the form $Bx + C$⁠) the general form would be

   $$\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4} ,$$

   with four unknowns. Because only $x^2$ appears, there is a shortcut: put $y = x^2$ in the fraction alone.

   This is not a substitution in the integral (we do not change $dx$⁠); it is just a name for $x^2$ while we do algebra. The book uses the same device in Example 14, Section 7.5. It works because no odd power of $x$ appears anywhere, so the $Ax$ and $Cx$ terms must turn out to be 0.
   Section 7.5, Example 14. Lesson: partial fractions

2. Split $\frac{1}{(y+1)(y+4)}$ into partial fractions

   $$\frac{1}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4} .$$

   Multiply by $(y+1)(y+4)$⁠:

   $$1 = A(y+4) + B(y+1) .$$

   $y = -1$⁠: $1 = 3A$⁠, so $A = \frac13$⁠.

   $y = -4$⁠: $1 = -3B$⁠, so $B = -\frac13$⁠.

   Confirm by coefficients: the coefficient of $y$ is $A + B = 0$ and the constant is $4A + B = \frac43 - \frac13 = 1$⁠, as required.

   The two factors $y+1$ and $y+4$ are different linear factors in $y$⁠, so Table 7.2, row 1 applies. The identity $1 = A(y+4) + B(y+1)$ holds for every $y$⁠, so we may put in $y = -1$ and $y = -4$ even though $y = x^2$ can never be negative: we are using an identity between polynomials in $y$⁠, not the value of $x^2$⁠.
   Table 7.2, row 1. Basics: splitting fractions

3. Put $x^2$ back

   $$\frac{1}{(x^2+1)(x^2+4)} = \frac13\left(\frac{1}{x^2+1} - \frac{1}{x^2+4}\right).$$

   Direct check: $\frac{1}{x^2+1} - \frac{1}{x^2+4}$ has numerator $(x^2+4) - (x^2+1) = 3$⁠, so the bracket is $\frac{3}{(x^2+1)(x^2+4)}$⁠, and one third of it is the integrand.

   The identity in $y$ holds for every $y \ne -1, -4$⁠, in particular for every value $y = x^2 \ge 0$⁠. The direct check confirms it without any reference to $y$⁠.
   Basics: splitting fractions

4. Integrate each piece

   $$\begin{aligned}&\int\frac{dx}{(x^2+1)(x^2+4)}\\ &= \frac13\int\frac{dx}{x^2+1^2} - \frac13\int\frac{dx}{x^2+2^2}\\ &= \frac13\tan^{-1}x - \frac13\cdot\frac12\tan^{-1}\frac{x}{2} + C\\ &= \frac13\tan^{-1}x - \frac16\tan^{-1}\frac{x}{2} + C .\end{aligned}$$
   Integral of a difference and constants outside (Properties (III), (IV), Section 7.2.1). The first piece is formula (3) of Section 7.4 with $a = 1$⁠; the second with $a = 2$⁠, which brings the factor $\frac12$ and the $\frac{x}{2}$ inside.
   Section 7.4, formula (3). Lesson: integrals of particular functions

5. Check: differentiate the answer

   Using $\frac{d}{dx}\tan^{-1}\frac{x}{2} = \frac{1}{1+\frac{x^2}{4}}\cdot\frac12 = \frac{2}{x^2+4}$⁠:

   $$\begin{aligned}&\frac{d}{dx}\left(\frac13\tan^{-1}x - \frac16\tan^{-1}\frac{x}{2}\right)\\ &= \frac{1}{3(x^2+1)} - \frac16\cdot\frac{2}{x^2+4}\\ &= \frac13\left(\frac{1}{x^2+1} - \frac{1}{x^2+4}\right)\\ &= \frac{1}{(x^2+1)(x^2+4)} .\ \checkmark\end{aligned}$$
   The last line is Step 3 read backwards. The derivative returns the integrand at every real $x$⁠, so the answer is valid on the whole real line.
   Chain rule for $\tan^{-1}\frac{x}{2}$⁠: inner function $\frac{x}{2}$⁠, derivative $\frac12$⁠. Basics: substitution

6. Compare with the textbook's answer

   The answer key gives $\frac13\tan^{-1}x - \frac16\tan^{-1}\frac{x}{2} + C$⁠, which is exactly our answer.

   Any correct answer may differ from the key by a constant; here the two forms are identical, so there is nothing to convert.

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