Exercise 7.4 · Q2: $\int\frac{dx}{\sqrt{1+4x^2}}$

Basics you need

Solution

1. Where is the integrand defined?

   $1 + 4x^2 \ge 1 > 0$ for every real $x$⁠, so the root is of a positive number and is never 0. The integrand is defined on the whole real line.

   The square root sits in a denominator, so we need the quantity under it to be strictly positive. That holds everywhere here, so the answer will need no restriction.

2. Substitute $t = 2x$

   Write $1 + 4x^2 = (2x)^2 + 1$ and put $t = 2x$⁠, so $dt = 2\,dx$⁠, that is $dx = \frac12\,dt$⁠:

   $$\int\frac{dx}{\sqrt{(2x)^2 + 1}} = \frac12\int\frac{dt}{\sqrt{t^2 + 1}}.$$
   The formula needs $t^2$ with coefficient 1 under the root. Absorbing the 4 into the variable ($4x^2 = (2x)^2$⁠) does that, and the price is the factor $\frac12$ from $dx = \frac12\,dt$⁠.
   Integration by substitution, Section 7.3.1. Lesson: integration by substitution

3. Use formula 7.4 (6) with $a = 1$

   $$\frac12\int\frac{dt}{\sqrt{t^2 + 1^2}} = \frac12\log\left|t + \sqrt{t^2 + 1}\right| + C.$$
   "Square plus a positive constant, under a root" is the shape of formula 7.4 (6). The book proves it by putting $x = a\tan\theta$⁠; with $-\frac\pi2 < \theta < \frac\pi2$ (where $\sec\theta > 0$⁠) this covers every real value of the variable.
   Formula 7.4 (6). Lesson: integrals of particular functions

4. Return to $x$

   With $t = 2x$⁠, $t^2 + 1 = 4x^2 + 1$⁠:

   $$\int\frac{dx}{\sqrt{1 + 4x^2}} = \frac12\log\left|2x + \sqrt{1 + 4x^2}\right| + C.$$

   The quantity inside the modulus is always positive, because $\sqrt{1 + 4x^2} > \sqrt{4x^2} = |2x| \ge -2x$⁠. So the modulus could be dropped, but keeping it (as the book does) is never wrong.

   A logarithm needs a positive input. Knowing that $2x + \sqrt{1 + 4x^2} > 0$ for every $x$ confirms the answer is defined everywhere the integrand is.

5. Check: differentiate the answer back

   Let $s = \sqrt{1 + 4x^2}$⁠. By Step 4, $2x + s > 0$⁠, so the answer is $F(x) = \frac12\log(2x + s)$ without the modulus. Then $s' = \frac{8x}{2s} = \frac{4x}{s}$⁠, and

   $$\begin{aligned} F'(x) &= \frac12\cdot\frac{2 + \frac{4x}{s}}{2x + s} \\ &= \frac12\cdot\frac{2(s + 2x)}{s\,(2x + s)} \\ &= \frac{1}{s} = \frac{1}{\sqrt{1 + 4x^2}}. \ \checkmark \end{aligned}$$
   Chain rule twice: $\frac{d}{dx}\log u = \frac{u'}{u}$ with $u = 2x + s$⁠, and $\frac{d}{dx}\sqrt{v} = \frac{v'}{2\sqrt v}$ with $v = 1 + 4x^2$⁠. Putting $2 + \frac{4x}{s}$ over the common denominator $s$ gives $\frac{2(s + 2x)}{s}$⁠, and the bracket $s + 2x$ cancels with the denominator $2x + s$⁠.
   Basics: integration undoes differentiation

6. Another route: take the 4 outside the root

   Alternatively, $\sqrt{1 + 4x^2} = \sqrt{4\left(x^2 + \frac14\right)} = 2\sqrt{x^2 + \left(\frac12\right)^2}$⁠, and formula 7.4 (6) with $a = \frac12$ gives

   $$\frac12\log\left|x + \sqrt{x^2 + \tfrac14}\right| + C'.$$

   This is the same family. Since $x + \sqrt{x^2 + \frac14} = \frac{2x + \sqrt{4x^2 + 1}}{2}$⁠,

   $$\begin{aligned} &\frac12\log\left|x + \sqrt{x^2 + \tfrac14}\right| \\ &= \frac12\log\left|2x + \sqrt{1 + 4x^2}\right| - \frac12\log 2. \end{aligned}$$
   The two answers differ by the constant $\frac12\log 2$⁠, which is absorbed into the arbitrary constant. Two correct methods may give different-looking answers, but they always differ by a constant (Property (II), Section 7.2.1).
   $\log\frac p2 = \log p - \log 2$⁠. Basics: logarithms

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