Example 8: $\int\frac{dx}{x^2 - 16}$ and $\int\frac{dx}{\sqrt{2x - x^2}}$

Basics you need

Part (i): $\int\frac{dx}{x^2 - 16}$

1. Recognise the shape $x^2 - a^2$

   $$\int\frac{dx}{x^2 - 16} = \int\frac{dx}{x^2 - 4^2}, \qquad a = 4.$$

   The integrand exists for $x \ne 4$ and $x \ne -4$⁠.

   16 is a perfect square, $4^2$⁠, so the denominator is a difference of squares, exactly the shape of formula 7.4 (1).
   Section 7.4. Lesson: integrals of particular functions

2. Apply formula 7.4 (1)

   $$\begin{aligned} \int\frac{dx}{x^2 - 4^2} &= \frac{1}{2\cdot 4}\log\left|\frac{x - 4}{x + 4}\right| + C \\ &= \frac18\log\left|\frac{x - 4}{x + 4}\right| + C. \end{aligned}$$
   With $a = 4$⁠, the factor $\frac{1}{2a}$ is $\frac18$⁠. Behind the formula: $\frac{1}{x^2 - 16} = \frac18\left(\frac{1}{x - 4} - \frac{1}{x + 4}\right)$⁠, and each piece integrates to a logarithm; the two logs combine into the log of a quotient.
   Formula 7.4 (1). Basics: splitting fractions

3. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\,\frac18\big(\log|x - 4| - \log|x + 4|\big) \\ &= \frac18\left(\frac{1}{x - 4} - \frac{1}{x + 4}\right) \\ &= \frac18\cdot\frac{(x + 4) - (x - 4)}{x^2 - 16} = \frac{1}{x^2 - 16}. \ \checkmark \end{aligned}$$

   Valid for every $x \ne \pm 4$⁠.

   Splitting the log of a quotient into a difference of logs makes the differentiation a two-term job; $(x + 4) - (x - 4) = 8$ cancels the $\frac18$⁠.

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

1. Complete the square under the root

   $$\begin{aligned} 2x - x^2 &= -(x^2 - 2x) \\ &= -\big[(x - 1)^2 - 1\big] \\ &= 1 - (x - 1)^2. \end{aligned}$$
   As written, $2x - x^2$ matches none of the six shapes. Completing the square writes it as "a constant minus a square", which is the $a^2 - x^2$ shape of formula 7.4 (5) with $a = 1$⁠.
   Basics: completing the square

2. Where the integrand exists

   We need $2x - x^2 > 0$⁠, that is $x(2 - x) > 0$⁠, so $0 < x < 2$⁠. Equivalently $(x - 1)^2 < 1$⁠, that is $-1 < x - 1 < 1$⁠.

   The square root must be of a positive number, and it is in a denominator, so it cannot be 0. This range is exactly where $\sin^{-1}(x - 1)$ is defined with a finite derivative, so the answer will be valid on the whole domain.

3. Shift: $t = x - 1$

   Put $t = x - 1$⁠; then $dt = dx$⁠, and $t$ runs over $(-1, 1)$⁠:

   $$\int\frac{dx}{\sqrt{1 - (x - 1)^2}} = \int\frac{dt}{\sqrt{1 - t^2}}.$$
   A shift by a constant does not change $dx$⁠, so the substitution only renames the inside of the square.
   Substitution, Section 7.3.1. Basics: substitution

4. Apply formula 7.4 (5) with $a = 1$ and return to $x$

   $$\begin{aligned} \int\frac{dt}{\sqrt{1 - t^2}} &= \sin^{-1}t + C \\ &= \sin^{-1}(x - 1) + C. \end{aligned}$$
   With $a = 1$⁠, formula 7.4 (5) is the same as standard formula (viii) of Section 7.2: $\frac{d}{dt}\sin^{-1}t = \frac{1}{\sqrt{1 - t^2}}$ for $-1 < t < 1$⁠.
   Formula 7.4 (5). Lesson: integrals of particular functions

5. Check: differentiate the answer

   $$\begin{aligned} \frac{d}{dx}\sin^{-1}(x - 1) &= \frac{1}{\sqrt{1 - (x - 1)^2}}\cdot 1 \\ &= \frac{1}{\sqrt{2x - x^2}}. \ \checkmark \end{aligned}$$

   Valid for $0 < x < 2$⁠.

   Chain rule with inner $x - 1$ (derivative 1); then Part (ii), Step 1 read backwards.

Every fact used, and where it comes from