Exercise 7.4 · Q19: $\int\frac{6x + 7}{\sqrt{(x - 5)(x - 4)}}\,dx$

Basics you need

Solution

1. Where the integrand exists

   We need $(x - 5)(x - 4) > 0$⁠: both factors positive or both negative, that is

   $$x < 4 \quad\text{or}\quad x > 5.$$
   The product is under a square root in a denominator, so it must be strictly positive. Between 4 and 5 the two factors have opposite signs. The domain has two separate pieces; the answer must work on both.

2. Expand, then split the numerator as $A\cdot Q' + B$

   $Q = (x - 5)(x - 4) = x^2 - 9x + 20$⁠, so $Q' = 2x - 9$⁠. Write

   $$6x + 7 = A(2x - 9) + B.$$

   Coefficients of $x$⁠: $2A = 6$⁠. Constants: $-9A + B = 7$⁠. So $A = 3$⁠, $B = 7 + 27 = 34$⁠.

   This is type 7.4 (10). Check: $3(2x - 9) + 34 = 6x - 27 + 34 = 6x + 7$⁠.
   Type 7.4 (10). Basics: splitting fractions

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

   $$\begin{aligned} \int\frac{6x + 7}{\sqrt Q}\,dx &= 3\int\frac{(2x - 9)\,dx}{\sqrt{x^2 - 9x + 20}} \\ &\quad + 34\int\frac{dx}{\sqrt{x^2 - 9x + 20}} \\ &= 3I_1 + 34I_2. \end{aligned}$$
   Dividing $3(2x - 9) + 34$ by $\sqrt Q$ 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 $t = x^2 - 9x + 20$ (so $t > 0$⁠), $dt = (2x - 9)\,dx$⁠:

   $$\begin{aligned} I_1 = \int\frac{dt}{\sqrt t} &= 2\sqrt t + C_1 \\ &= 2\sqrt{x^2 - 9x + 20} + C_1. \end{aligned}$$
   Power formula with $n = -\frac12$⁠: $\int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} = 2\sqrt t$⁠.
   Substitution, Section 7.3.1; standard formula (i), Section 7.2. Basics: the power rule

5. $I_2$ by completing the square

   $$\begin{aligned} x^2 - 9x + 20 &= \left(x - \frac92\right)^2 - \frac{81}{4} + 20 \\ &= \left(x - \frac92\right)^2 - \left(\frac12\right)^2. \end{aligned}$$

   Put $t = x - \frac92$⁠, $dt = dx$⁠; on the domain $|t| > \frac12$⁠. By 7.4 (4) with $a = \frac12$⁠:

   $$\begin{aligned} I_2 &= \int\frac{dt}{\sqrt{t^2 - \left(\frac12\right)^2}} = \log\left|t + \sqrt{t^2 - \tfrac14}\right| + C_2 \\ &= \log\left|x - \frac92 + \sqrt{x^2 - 9x + 20}\right| + C_2. \end{aligned}$$
   Half of $-9$ is $-\frac92$⁠, and $20 - \frac{81}{4} = -\frac14$⁠, a negative leftover, so this is "square minus square". The book proves 7.4 (4) with $x = a\sec\theta$⁠; the result also holds for $t < -\frac12$⁠, because the derivative of $\log\left|t + \sqrt{t^2 - a^2}\right|$ is $\frac{1}{\sqrt{t^2 - a^2}}$ whenever $|t| > a$ (shown in Step 7). So it covers both pieces of the domain.
   Type 7.4 (8); formula 7.4 (4). Basics: completing the square

6. Combine

   $$\begin{aligned} \int\frac{6x + 7}{\sqrt{(x - 5)(x - 4)}}\,dx &= 6\sqrt{x^2 - 9x + 20} \\ &\quad + 34\log\left|x - \frac92 + \sqrt{x^2 - 9x + 20}\right| \\ &\quad + C, \end{aligned}$$

   with $C = 3C_1 + 34C_2$⁠.

   $3\cdot 2\sqrt Q = 6\sqrt Q$⁠. The constants merge into one. This is exactly the textbook's form.

7. Check: differentiate the answer

   Write $s = \sqrt{x^2 - 9x + 20}$ and $t = x - \frac92$⁠, so $s^2 = t^2 - \frac14$ and $s' = \frac{2x - 9}{2s} = \frac ts$⁠. Then

   $$\begin{aligned} \frac{d}{dx}\log|t + s| &= \frac{1 + \frac ts}{t + s} = \frac1s, \\ \frac{d}{dx}\left(6s\right) &= \frac{3(2x - 9)}{s} = \frac{6x - 27}{s}. \end{aligned}$$

   So the derivative of our answer is

   $$\frac{6x - 27}{s} + \frac{34}{s} = \frac{6x + 7}{\sqrt{(x - 5)(x - 4)}}. \ \checkmark$$
   In the first line, $1 + \frac ts = \frac{s + t}{s}$⁠, and the factor $t + s$ cancels. It is never 0: $s^2 < t^2$⁠, so $s < |t|$ and $t + s$ has the sign of $t$⁠. Nothing here assumed $x > 5$⁠, so the check holds on both pieces of the domain.
   Chain rule (Theorem 4, Section 5.3.1); derivative (xii), Section 7.2. Basics: logarithms and 1/x

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