Exercise 7.7 · Q3: $\int\sqrt{x^2 + 4x + 6}\,dx$

Basics you need

Solution

1. Complete the square

   Half the coefficient of $x$ is $\frac42 = 2$⁠, and $2^2 = 4$⁠. Add and subtract 4:

   $$\begin{aligned} x^2 + 4x + 6 &= (x^2 + 4x + 4) + 6 - 4 \\ &= (x + 2)^2 + 2. \end{aligned}$$
   $(x + 2)^2 = x^2 + 4x + 4$⁠, so the first three terms form a perfect square and the leftover is $6 - 4 = 2$⁠. The sign of the leftover decides the formula: positive leftover with $+y^2$ means 7.6.2 (ii). This is the method of Section 7.4 (7) and Example 23.
   Completing the square. Basics: completing the square

2. The domain: every real $x$

   Since $(x + 2)^2 \ge 0$⁠, we get $x^2 + 4x + 6 = (x + 2)^2 + 2 \ge 2 > 0$ for every real $x$⁠. So the integrand is defined (and positive) for all $x$⁠.

   A square root needs a non-negative inside. Completing the square shows this at a glance: the smallest value of the quadratic is 2, at $x = -2$⁠.

3. Put $y = x + 2$

   Then $dy = dx$ and

   $$\int\sqrt{x^2 + 4x + 6}\,dx = \int\sqrt{y^2 + \left(\sqrt2\right)^2}\,dy.$$
   A shift $y = x + 2$ has $\frac{dy}{dx} = 1$⁠, so $dx$ becomes $dy$ with no extra factor. Now the root has the shape $\sqrt{y^2 + a^2}$ with $a = \sqrt2$⁠.
   Integration by substitution, Section 7.3.1. Basics: substitution

4. Where formula 7.6.2 (ii) comes from

   Let $a > 0$ and $I = \int\sqrt{y^2 + a^2}\cdot 1\,dy$⁠. By parts with first function $\sqrt{y^2 + a^2}$ and second function 1:

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

   By 7.4 (6) the last integral is $\log\left|y + \sqrt{y^2 + a^2}\right|$⁠. So $2I = y\sqrt{y^2 + a^2}$ $+ a^2\log\left|y + \sqrt{y^2 + a^2}\right|$⁠, and

   $$\begin{aligned} I &= \frac{y}{2}\sqrt{y^2 + a^2} \\ &\quad + \frac{a^2}{2}\log\left|y + \sqrt{y^2 + a^2}\right| + C. \end{aligned}$$
   The derivative of $\sqrt{y^2 + a^2}$ is $\frac{y}{\sqrt{y^2 + a^2}}$⁠. Writing $y^2 = (y^2 + a^2) - a^2$ makes the original integral $I$ reappear, and we solve for it. The textbook does exactly this for formula (i) in Section 7.6.2 and says (ii) follows "similarly".
   Integration by parts, Section 7.6; formula 7.4 (6). Lesson: integration by parts

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

   With $a^2 = 2$⁠, $\frac{a^2}{2} = 1$⁠:

   $$\begin{aligned} \int\sqrt{y^2 + 2}\,dy &= \frac{y}{2}\sqrt{y^2 + 2} \\ &\quad + \log\left|y + \sqrt{y^2 + 2}\right| + C. \end{aligned}$$

   Now $y = x + 2$ and $y^2 + 2 = x^2 + 4x + 6$⁠, so

   $$\begin{aligned} \int\sqrt{x^2 + 4x + 6}\,dx &= \frac{x + 2}{2}\sqrt{x^2 + 4x + 6} \\ &\quad + \log\left|x + 2 + \sqrt{x^2 + 4x + 6}\right| + C. \end{aligned}$$
   Undo the substitution everywhere: $y^2 + 2$ is the completed square from Step 1, which is $x^2 + 4x + 6$ again. Note that $x + 2 + \sqrt{x^2 + 4x + 6}$ is always positive (the root $\sqrt{(x + 2)^2 + 2}$ is bigger than $|x + 2|$⁠), so the modulus could be dropped here; the textbook keeps it, and it does no harm.
   Formula 7.6.2 (ii). Lesson: integration by parts

6. Check: differentiate the answer

   Write $y = x + 2$ and $r = \sqrt{y^2 + 2}$⁠, so $\frac{dy}{dx} = 1$ and $\frac{dr}{dx} = \frac{y}{r}$⁠. Then

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

   Adding, over the common denominator $2r$⁠:

   $$\begin{aligned} \frac{r^2 + y^2 + 2}{2r} &= \frac{2(y^2 + 2)}{2r} = \frac{r^2}{r} \\ &= r = \sqrt{x^2 + 4x + 6}.\ \checkmark \end{aligned}$$
   $1 + \frac{y}{r} = \frac{r + y}{r}$⁠, and dividing by $y + r$ leaves $\frac1r$⁠. Then $\frac{r}{2} = \frac{r^2}{2r}$ with $r^2 = y^2 + 2$⁠, so the numerator is $2(y^2 + 2) = 2r^2$⁠. This works for every real $x$⁠.
   Product rule; chain rule; standard formula (xii), Section 7.2. Basics: logarithms and 1/x

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