Example 23: $\int\sqrt{x^2+2x+5}\,dx$

Basics you need

Solution

1. Complete the square under the root

   Here $b = 2$⁠, so $\frac{b}{2} = 1$ and $\frac{b^2}{4} = 1$⁠. Split the 5 as $1 + 4$⁠:

   $$x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2.$$

   So the integral is $\int\sqrt{(x+1)^2 + 2^2}\,dx$⁠. Since a square is never negative, $(x+1)^2 + 4 \ge 4$⁠: the root is defined, and positive, for every real $x$⁠.

   None of the three standard formulas has a middle term like $2x$⁠. Completing the square hides that term inside a single square $(x+1)^2$⁠, and what is left over, $+4$⁠, is a positive constant, which we write as $a^2$ with $a = 2$⁠. The sign of the leftover decides the formula: $+a^2$ here, so formula (ii).
   Check the square by expanding: $(x+1)^2 + 4 = x^2 + 2x + 1 + 4 = x^2 + 2x + 5$⁠. Basics: completing the square

2. Shift: put $y = x + 1$

   Let $y = x + 1$⁠. Then $\frac{dy}{dx} = 1$⁠, so $dy = dx$⁠, and

   $$\int\sqrt{x^2 + 2x + 5}\,dx = \int\sqrt{y^2 + 2^2}\,dy.$$
   The new variable $y$ absorbs the "$+1$" so that the integrand looks exactly like the standard one. Because the derivative of $x + 1$ is 1, the differential needs no correction factor: $dx$ simply becomes $dy$⁠.
   Integration by substitution (Section 7.3.1): if $y = g(x)$ then $dy = g'(x)\,dx$⁠. Here $g'(x) = 1$⁠. Lesson: integration by substitution

3. Apply formula 7.6.2 (ii) with $a = 2$

   With $a = 2$ we have $a^2 = 4$ and $\frac{a^2}{2} = 2$⁠. Formula (ii), written in the letter $y$⁠, gives

   $$\int\sqrt{y^2 + 4}\,dy = \frac{1}{2}y\sqrt{y^2 + 4} + 2\log\left|y + \sqrt{y^2 + 4}\right| + C.$$
   The formula holds for any letter used as the variable, so we may apply it in $y$⁠. The only care needed is the coefficient of the log: it is $\frac{a^2}{2} = \frac{4}{2} = 2$⁠, not $a^2 = 4$⁠. It happens to equal $a$ here, but only because $a = 2$⁠; in general always compute $\frac{a^2}{2}$⁠.
   Formula 7.6.2 (ii) is the one with $+a^2$ inside the root and a log in the answer; the one with $a^2 - x^2$ has $\sin^{-1}$ instead (used in Example 24). Lesson: integration by parts

4. Return to $x$

   Replace $y$ by $x + 1$ and $y^2 + 4$ by $x^2 + 2x + 5$ (Step 1):

   $$\begin{aligned}\int\sqrt{x^2 + 2x + 5}\,dx = {}&\tfrac{1}{2}(x + 1)\sqrt{x^2 + 2x + 5} \\ &+ 2\log\left|x + 1 + \sqrt{x^2 + 2x + 5}\right| \\ &+ C.\end{aligned}$$

   Where it is valid. The root is defined for all real $x$ (Step 1). The quantity inside the log is never zero; it is in fact always positive, because $\sqrt{y^2 + 4} > \sqrt{y^2} = |y| \ge -y$⁠, so $y + \sqrt{y^2 + 4} > 0$⁠. Hence the answer holds for every real $x$⁠, and the bars $|\;|$ could even be dropped.

   The question was asked in $x$⁠, so the answer must be in $x$⁠. Writing $y^2 + 4$ back as $x^2 + 2x + 5$ (rather than $(x+1)^2 + 4$⁠) matches the book's printed form.
   $\log|u|$ needs $u \ne 0$⁠; a quantity that is always positive satisfies this everywhere. Basics: logarithms and 1/x

5. Check: differentiate the answer

   Write $y = x + 1$ and $R = \sqrt{y^2 + 4}$⁠, so that $\frac{dy}{dx} = 1$ and, by the chain rule, $\frac{dR}{dx} = \frac{2y}{2\sqrt{y^2+4}} = \frac{y}{R}$⁠.

   First term, by the product rule:

   $$\frac{d}{dx}\left(\tfrac{1}{2}yR\right) = \tfrac{1}{2}R + \tfrac{1}{2}y\cdot\frac{y}{R} = \frac{R^2 + y^2}{2R}.$$

   Second term, by the chain rule for $\log$⁠:

   $$\frac{d}{dx}\,2\log(y + R) = \frac{2\left(1 + \frac{y}{R}\right)}{y + R} = \frac{2\cdot\frac{R + y}{R}}{y + R} = \frac{2}{R}.$$

   Add them, using $R^2 = y^2 + 4$⁠:

   $$\frac{R^2 + y^2 + 4}{2R} = \frac{2y^2 + 8}{2R} = \frac{y^2 + 4}{R} = R.$$

   And $R = \sqrt{(x+1)^2 + 4} = \sqrt{x^2 + 2x + 5}$⁠, the integrand. The constant C differentiates to 0. The answer is confirmed.

   An antiderivative is defined by the property that its derivative is the integrand, so differentiating is the complete check. It also proves formula 7.6.2 (ii) in this case without the integration by parts.
   Integration undoes differentiation: $\int f(x)\,dx = F(x) + C$ means exactly $F'(x) = f(x)$⁠. Basics: integration undoes differentiation

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