Example 9: $\int\frac{dx}{x^2 - 6x + 13}$⁠, $\int\frac{dx}{3x^2 + 13x - 10}$⁠, $\int\frac{dx}{\sqrt{5x^2 - 2x}}$

Basics you need

Part (i): $\int\frac{dx}{x^2 - 6x + 13}$

1. Complete the square

   $$\begin{aligned} x^2 - 6x + 13 &= x^2 - 6x + 9 - 9 + 13 \\ &= (x - 3)^2 + 4 \\ &= (x - 3)^2 + 2^2. \end{aligned}$$
   Half the coefficient of $x$ is $-3$⁠; adding and subtracting $3^2 = 9$ creates the perfect square $(x - 3)^2$ without changing the value. The leftover $+4$ is positive, so the denominator is never 0 and the integrand exists for every real $x$⁠.
   Basics: completing the square

2. Shift $t = x - 3$ and use 7.4 (3)

   Put $t = x - 3$⁠, $dt = dx$⁠:

   $$\begin{aligned} \int\frac{dt}{t^2 + 2^2} &= \frac12\tan^{-1}\frac t2 + C \\ &= \frac12\tan^{-1}\frac{x - 3}{2} + C. \end{aligned}$$
   "Square plus a positive constant" is the shape $t^2 + a^2$ with $a = 2$⁠, so formula 7.4 (3) applies, giving $\frac1a\tan^{-1}\frac ta$⁠.
   Formula 7.4 (3). Lesson: integrals of particular functions

3. Check: differentiate the answer

   $$\begin{aligned} \frac{d}{dx}\,\frac12\tan^{-1}\frac{x - 3}{2} &= \frac12\cdot\frac{1}{1 + \frac{(x - 3)^2}{4}}\cdot\frac12 \\ &= \frac{1}{4 + (x - 3)^2} = \frac{1}{x^2 - 6x + 13}. \ \checkmark \end{aligned}$$
   Chain rule: $\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2}$ with $u = \frac{x - 3}{2}$⁠, $u' = \frac12$⁠. Multiplying top and bottom by 4 clears the inner fraction.

Part (ii): $\int\frac{dx}{3x^2 + 13x - 10}$

1. Take out the 3 and complete the square

   $$\begin{aligned} 3x^2 + 13x - 10 &= 3\left(x^2 + \frac{13}{3}x - \frac{10}{3}\right) \\ &= 3\left[\left(x + \frac{13}{6}\right)^2 - \frac{169}{36} - \frac{120}{36}\right] \\ &= 3\left[\left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2\right]. \end{aligned}$$
   Half of $\frac{13}{3}$ is $\frac{13}{6}$⁠, whose square is $\frac{169}{36}$⁠; and $\frac{10}{3} = \frac{120}{36}$⁠. Their sum is $\frac{289}{36} = \left(\frac{17}{6}\right)^2$⁠. The leftover is negative, so this is the "square minus square" shape: a logarithm is coming. (Indeed $3x^2 + 13x - 10 = (3x - 2)(x + 5)$⁠, so the integrand needs $x \ne \frac23$ and $x \ne -5$⁠.)
   Basics: completing the square

2. Shift $t = x + \frac{13}{6}$ and use 7.4 (1)

   $$\begin{aligned} \int\frac{dx}{3x^2 + 13x - 10} &= \frac13\int\frac{dt}{t^2 - \left(\frac{17}{6}\right)^2} \\ &= \frac13\cdot\frac{1}{2\cdot\frac{17}{6}}\log\left|\frac{t - \frac{17}{6}}{t + \frac{17}{6}}\right| + C_1 \\ &= \frac{1}{17}\log\left|\frac{t - \frac{17}{6}}{t + \frac{17}{6}}\right| + C_1. \end{aligned}$$
   Formula 7.4 (1) with $a = \frac{17}{6}$⁠: $\frac{1}{2a} = \frac{6}{34} = \frac{3}{17}$⁠, and $\frac13\cdot\frac{3}{17} = \frac{1}{17}$⁠. (The book's margin note "[by 7.4 (i)]" refers to this formula, printed elsewhere as 7.4 (1).)
   Formula 7.4 (1). Lesson: integrals of particular functions

3. Return to $x$ and simplify

   $t - \frac{17}{6} = x - \frac46 = \frac{6x - 4}{6}$ and $t + \frac{17}{6} = x + \frac{30}{6} = \frac{6x + 30}{6}$⁠, so

   $$\frac{t - \frac{17}{6}}{t + \frac{17}{6}} = \frac{6x - 4}{6x + 30} = \frac13\cdot\frac{3x - 2}{x + 5}.$$

   Then

   $$\begin{aligned} &\frac{1}{17}\log\left|\frac13\cdot\frac{3x - 2}{x + 5}\right| + C_1 \\ &= \frac{1}{17}\log\left|\frac{3x - 2}{x + 5}\right| + C_1 + \frac{1}{17}\log\frac13 \\ &= \frac{1}{17}\log\left|\frac{3x - 2}{x + 5}\right| + C, \end{aligned}$$

   where $C = C_1 + \frac{1}{17}\log\frac13$⁠.

   $6x - 4 = 2(3x - 2)$ and $6x + 30 = 6(x + 5)$⁠, so the quotient carries a constant factor $\frac26 = \frac13$⁠. The log of a product is a sum of logs, and $\frac{1}{17}\log\frac13$ is a constant, which joins $C_1$⁠.
   $\log|kp| = \log|k| + \log|p|$⁠. Basics: logarithms

4. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\,\frac{1}{17}\big(\log|3x - 2| - \log|x + 5|\big) \\ &= \frac{1}{17}\left(\frac{3}{3x - 2} - \frac{1}{x + 5}\right) \\ &= \frac{1}{17}\cdot\frac{3x + 15 - 3x + 2}{(3x - 2)(x + 5)} \\ &= \frac{1}{(3x - 2)(x + 5)} \\ &= \frac{1}{3x^2 + 13x - 10}. \ \checkmark \end{aligned}$$
   $\frac{d}{dx}\log|3x - 2| = \frac{3}{3x - 2}$ by the chain rule. The numerator collapses to 17, which cancels the $\frac{1}{17}$⁠. Valid for $x \ne \frac23, -5$⁠.

Part (iii): $\int\frac{dx}{\sqrt{5x^2 - 2x}}$

1. Where the integrand exists

   We need $5x^2 - 2x = x(5x - 2) > 0$⁠: both factors positive ($x > \frac25$⁠) or both negative ($x < 0$⁠).

   The root must be of a positive number because it sits in a denominator. The domain has two separate pieces; the answer must work on both.

2. Take out the 5 and complete the square

   $$\begin{aligned} 5x^2 - 2x &= 5\left(x^2 - \frac{2x}{5}\right) \\ &= 5\left[\left(x - \frac15\right)^2 - \left(\frac15\right)^2\right]. \end{aligned}$$

   So $\sqrt{5x^2 - 2x} = \sqrt5\,\sqrt{\left(x - \frac15\right)^2 - \left(\frac15\right)^2}$⁠.

   Half of $-\frac25$ is $-\frac15$⁠. The square root of a product of positive numbers is the product of the square roots, which lets the $\sqrt5$ come outside.
   Basics: completing the square

3. Shift $t = x - \frac15$ and use 7.4 (4)

   Put $t = x - \frac15$⁠, $dt = dx$⁠. On the domain, $t^2 > \left(\frac15\right)^2$⁠:

   $$\begin{aligned} \int\frac{dx}{\sqrt{5x^2 - 2x}} &= \frac{1}{\sqrt5}\int\frac{dt}{\sqrt{t^2 - \left(\frac15\right)^2}} \\ &= \frac{1}{\sqrt5}\log\left|t + \sqrt{t^2 - \left(\frac15\right)^2}\right| + C. \end{aligned}$$
   "Square minus square" under a root is the shape of formula 7.4 (4), here with $a = \frac15$⁠. The book proves 7.4 (4) with $x = a\sec\theta$⁠; the result also holds for $t < -a$⁠, because $\frac{d}{dt}\log\left|t + \sqrt{t^2 - a^2}\right| = \frac{1}{\sqrt{t^2 - a^2}}$ whenever $|t| > a$ (shown in Part (iii), Step 5). So it covers both pieces of the domain.
   Formula 7.4 (4). Lesson: integrals of particular functions

4. Return to $x$

   $t^2 - \frac{1}{25}$ $= x^2 - \frac{2x}{5} + \frac{1}{25} - \frac{1}{25}$ $= x^2 - \frac{2x}{5}$⁠, so

   $$\begin{aligned} &\int\frac{dx}{\sqrt{5x^2 - 2x}} \\ &=\frac{1}{\sqrt5}\log\left|x - \frac15 + \sqrt{x^2 - \frac{2x}{5}}\right| + C. \end{aligned}$$
   Expanding $\left(x - \frac15\right)^2$ brings back the original $x^2 - \frac{2x}{5}$⁠, so the answer can be written without $t$⁠. This is the book's final form.

5. Check: differentiate the answer

   Write $s = \sqrt{x^2 - \frac{2x}{5}}$⁠, so $s' = \frac{x - \frac15}{s}$⁠. Then

   $$\begin{aligned} \frac{d}{dx}\log\left|x - \tfrac15 + s\right| &= \frac{1 + \frac{x - \frac15}{s}}{x - \frac15 + s} \\ &= \frac{\frac{s + x - \frac15}{s}}{x - \frac15 + s} = \frac1s. \end{aligned}$$

   So the derivative of our answer is $\frac{1}{\sqrt5\,s} = \frac{1}{\sqrt{5x^2 - 2x}}$⁠. $\checkmark$

   The numerator and the denominator contain the same sum $x - \frac15 + s$⁠, which cancels; it is never 0 because $s < \left|x - \frac15\right|$⁠. Nothing here assumed $x > \frac25$⁠, so the check holds on both pieces of the domain. Finally $\sqrt5\cdot s = \sqrt{5x^2 - 2x}$⁠.

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