Exercise 7.4 · Q22: $\int\frac{x + 3}{x^2 - 2x - 5}\,dx$

Basics you need

Solution

1. The domain

   The denominator vanishes where $x^2 - 2x - 5 = 0$⁠, that is

   $$x = \frac{2 \pm \sqrt{4 + 20}}{2} = 1 \pm \sqrt6.$$

   So the integrand is defined for every real $x \ne 1 \pm \sqrt6$ (roughly $x \ne -1.449$ and $x \ne 3.449$⁠). The domain has three pieces.

   A fraction is undefined where its denominator is 0. The answer below must work on each of the three intervals $x < 1 - \sqrt6$⁠, $1 - \sqrt6 < x < 1 + \sqrt6$⁠, $x > 1 + \sqrt6$⁠.

2. Split the numerator as $A\cdot Q' + B$

   $Q = x^2 - 2x - 5$ has $Q' = 2x - 2$⁠. Write

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

   Coefficients of $x$⁠: $2A = 1$⁠. Constants: $-2A + B = 3$⁠. So $A = \frac12$⁠, $B = 3 + 1 = 4$⁠.

   This is type 7.4 (9). Check: $\frac12(2x - 2) + 4 = x - 1 + 4 = x + 3$⁠.
   Type 7.4 (9). Basics: splitting fractions

3. Split the integral into $I_1$ and $I_2$

   $$\begin{aligned} \int\frac{x + 3}{x^2 - 2x - 5}\,dx &= \frac12\int\frac{2x - 2}{x^2 - 2x - 5}\,dx \\ &\quad + 4\int\frac{dx}{x^2 - 2x - 5} \\ &= \frac12 I_1 + 4I_2. \end{aligned}$$
   Dividing $\frac12(2x - 2) + 4$ by $Q$ term by term gives two fractions; Property (V) of Section 7.2.1 splits the integral.
   Property (V), Section 7.2.1. Lesson: integrals of particular functions

4. $I_1$ by substitution

   Put $t = x^2 - 2x - 5$⁠, $dt = (2x - 2)\,dx$⁠:

   $$I_1 = \int\frac{dt}{t} = \log|t| + C_1 = \log\left|x^2 - 2x - 5\right| + C_1.$$
   The numerator $2x - 2$ is exactly the derivative of the denominator. Here $Q$ does change sign (it is negative between its roots), so the modulus is essential.
   Substitution, Section 7.3.1; standard formula (xii), Section 7.2. Basics: substitution

5. $I_2$ by completing the square

   $x^2 - 2x - 5 = (x - 1)^2 - 6$ $= (x - 1)^2 - \left(\sqrt6\right)^2$⁠. Put $t = x - 1$⁠, $dt = dx$⁠, and use 7.4 (1) with $a = \sqrt6$⁠:

   $$\begin{aligned} I_2 &= \int\frac{dt}{t^2 - \left(\sqrt6\right)^2} = \frac{1}{2\sqrt6}\log\left|\frac{t - \sqrt6}{t + \sqrt6}\right| + C_2 \\ &= \frac{1}{2\sqrt6}\log\left|\frac{x - 1 - \sqrt6}{x - 1 + \sqrt6}\right| + C_2. \end{aligned}$$
   Half of $-2$ is $-1$⁠, and $-5 - 1 = -6$⁠, a negative leftover, so this is "square minus square" without a root: formula 7.4 (1). It holds wherever $t \ne \pm\sqrt6$⁠, that is, on all three pieces of the domain.
   Type 7.4 (7); formula 7.4 (1). Basics: completing the square

6. Combine

   $$\begin{aligned} \int\frac{x + 3}{x^2 - 2x - 5}\,dx &= \frac12\log\left|x^2 - 2x - 5\right| \\ &\quad + \frac{2}{\sqrt6}\log\left|\frac{x - 1 - \sqrt6}{x - 1 + \sqrt6}\right| + C, \end{aligned}$$

   with $C = \frac{C_1}{2} + 4C_2$⁠.

   $4\cdot\frac{1}{2\sqrt6} = \frac{2}{\sqrt6}$⁠. The constants merge into one. This is exactly the textbook's form.

7. Check: differentiate the answer

   Write $r = \sqrt6$⁠, so $Q = (x - 1)^2 - r^2 = (x - 1 - r)(x - 1 + r)$⁠. Then

   $$\begin{aligned} \frac{d}{dx}\left[\frac12\log|Q|\right] &= \frac{2x - 2}{2Q} = \frac{x - 1}{Q}, \\ \frac{d}{dx}\log\left|\frac{x - 1 - r}{x - 1 + r}\right| &= \frac{1}{x - 1 - r} - \frac{1}{x - 1 + r} \\ &= \frac{2r}{Q}. \end{aligned}$$

   So the derivative of our answer is $\frac{x - 1}{Q} + \frac{2}{r}\cdot\frac{2r}{Q} = \frac{x - 1 + 4}{Q} = \frac{x + 3}{x^2 - 2x - 5}$⁠. $\checkmark$

   The log of a ratio is the difference of the logs, $\log|x - 1 - r| - \log|x - 1 + r|$⁠, and each differentiates to 1 over its inside. Over the common denominator $Q$⁠, the difference of the two fractions is $\frac{(x - 1 + r) - (x - 1 - r)}{Q} = \frac{2r}{Q}$⁠. Valid for every $x \ne 1 \pm \sqrt6$⁠.
   Laws of logarithms; derivative (xii), Section 7.2. Basics: logarithms and 1/x

Every fact used, and where it comes from