Exercise 7.7 · Q4: $\int\sqrt{x^2 + 4x + 1}\,dx$

Basics you need

Solution

1. Complete the square

   Half the coefficient of $x$ is 2, and $2^2 = 4$⁠:

   $$\begin{aligned} x^2 + 4x + 1 &= (x^2 + 4x + 4) + 1 - 4 \\ &= (x + 2)^2 - 3. \end{aligned}$$
   The leftover $1 - 4 = -3$ is negative, so the shape is "square minus $a^2$" with $a = \sqrt3$⁠: formula 7.6.2 (i).
   Completing the square, as in Section 7.4 (7). Basics: completing the square

2. The domain has two pieces

   We need $(x + 2)^2 - 3 \ge 0$⁠, that is $|x + 2| \ge \sqrt3$⁠:

   $$x \le -2 - \sqrt3 \quad\text{or}\quad x \ge -2 + \sqrt3.$$

   Numerically, $x \le -3.732$ or $x \ge -0.268$ (to three decimals). Between these values the quadratic is negative and the integrand does not exist.

   $|x + 2| \ge \sqrt3$ means $x + 2 \ge \sqrt3$ or $x + 2 \le -\sqrt3$⁠. The roots of $x^2 + 4x + 1 = 0$ are $-2 \pm \sqrt3$⁠; the upward parabola is non-negative outside them. Because the domain has a gap, the constant $C$ may even be chosen differently on the two pieces.

3. Put $y = x + 2$

   Then $dy = dx$ and

   $$\int\sqrt{x^2 + 4x + 1}\,dx = \int\sqrt{y^2 - \left(\sqrt3\right)^2}\,dy.$$
   A shift has derivative 1, so $dx = dy$⁠. The domain becomes $|y| \ge \sqrt3$⁠, exactly where $\sqrt{y^2 - 3}$ is real.
   Integration by substitution, Section 7.3.1. Basics: substitution

4. How formula 7.6.2 (i) is derived

   For $a > 0$⁠, $|y| > a$⁠, let $I = \int\sqrt{y^2 - a^2}\cdot 1\,dy$⁠. By parts (first function the root, second function 1), and writing $y^2 = (y^2 - a^2) + a^2$⁠:

   $$\begin{aligned} I &= y\sqrt{y^2 - a^2} - \int\frac{y^2\,dy}{\sqrt{y^2 - a^2}} \\ &= y\sqrt{y^2 - a^2} - I - a^2\int\frac{dy}{\sqrt{y^2 - a^2}}. \end{aligned}$$

   With 7.4 (4) and solving $2I = \dots$ for $I$⁠:

   $$\begin{aligned} I &= \frac{y}{2}\sqrt{y^2 - a^2} \\ &\quad - \frac{a^2}{2}\log\left|y + \sqrt{y^2 - a^2}\right| + C. \end{aligned}$$
   This is the textbook's own derivation in Section 7.6.2. The derivative of $\sqrt{y^2 - a^2}$ is $\frac{y}{\sqrt{y^2 - a^2}}$⁠, so the product rule term is $y\cdot\frac{y}{\sqrt{y^2 - a^2}}$⁠. The $a^2$ part comes with a minus sign this time, which is why the log is subtracted.
   Integration by parts, Section 7.6; formula 7.4 (4). Lesson: integration by parts

5. Apply it with $a^2 = 3$ and return to $x$

   With $a^2 = 3$⁠, $\frac{a^2}{2} = \frac32$⁠. Putting back $y = x + 2$ and $y^2 - 3 = x^2 + 4x + 1$⁠:

   $$\begin{aligned} \int\sqrt{x^2 + 4x + 1}\,dx &= \frac{x + 2}{2}\sqrt{x^2 + 4x + 1} \\ &\quad - \frac32\log\left|x + 2 + \sqrt{x^2 + 4x + 1}\right| + C. \end{aligned}$$
   Here the modulus really matters. On the left piece, $x + 2 \le -\sqrt3$ is negative and $\sqrt{x^2 + 4x + 1} < |x + 2|$⁠, so $x + 2 + \sqrt{x^2 + 4x + 1}$ is negative there. It is never 0: that would need $\sqrt{(x + 2)^2 - 3} = -(x + 2)$⁠, and squaring gives $-3 = 0$⁠, impossible. So the log is defined on the whole domain.
   Formula 7.6.2 (i). Lesson: integration by parts

6. Check: differentiate the answer

   Write $y = x + 2$⁠, $r = \sqrt{y^2 - 3}$⁠, so $\frac{dr}{dx} = \frac{y}{r}$ where $r > 0$⁠. Then

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

   So the derivative of our answer is, over the denominator $2r$⁠,

   $$\begin{aligned} \frac{r^2 + y^2 - 3}{2r} &= \frac{2(y^2 - 3)}{2r} = \frac{r^2}{r} \\ &= r = \sqrt{x^2 + 4x + 1}.\ \checkmark \end{aligned}$$
   $-\frac32\cdot\frac1r = -\frac{3}{2r}$⁠, and $\frac r2 = \frac{r^2}{2r}$ with $r^2 = y^2 - 3$⁠. The rule $\frac{d}{dx}\log|u| = \frac{u'}{u}$ holds for negative $u$ as well, so the check covers both pieces. It needs $r > 0$⁠, so it applies at every point of the domain except the two end points $x = -2 \pm \sqrt3$⁠; there the integrand is 0 and the answer's one-sided slope is 0 as well.
   Product rule; chain rule; standard formula (xii), Section 7.2. Basics: logarithms and 1/x

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