Exercise 7.1 · Q10: $\int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 dx$

Basics you need

Solution

1. Fix the domain

   $\sqrt{x}$ needs $x \ge 0$⁠, and $\frac{1}{\sqrt{x}}$ also needs $\sqrt{x} \ne 0$⁠. So the integrand is defined exactly for $x > 0$⁠, and we work on the interval $(0, \infty)$⁠.

   Expanding will produce $\frac1x$⁠, which looks as if it allowed negative $x$ too. It does not: the original integrand never existed there. Recording the domain now keeps the answer honest.
   Note after the standard formulae, Section 7.2 (keep the interval in mind). Basics: powers and roots

2. Expand the square

   With $p = \sqrt{x}$ and $q = \frac{1}{\sqrt{x}}$⁠:

   $$\begin{aligned} p^2 &= \left(\sqrt{x}\right)^2 = x, \\ 2pq &= 2\sqrt{x}\cdot\frac{1}{\sqrt{x}} = 2, \\ q^2 &= \frac{1}{\left(\sqrt{x}\right)^2} = \frac{1}{x}. \end{aligned}$$

   Hence, for $x > 0$⁠,

   $$\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 = x - 2 + \frac{1}{x}.$$
   An integral of a square has no shortcut of its own, but a sum of powers does. The middle term simplifies to the constant $2$ because $\sqrt{x}$ cancels with $\frac{1}{\sqrt{x}}$ (allowed since $\sqrt{x} \ne 0$⁠).
   $(p - q)^2 = p^2 - 2pq + q^2$⁠; $\left(\sqrt{x}\right)^2 = x$ for $x \ge 0$⁠. Basics: splitting fractions

3. Integrate term by term

   $$\begin{aligned} &\int\left(x - 2 + \frac1x\right)dx \\ &= \int x\,dx - 2\int dx + \int\frac1x\,dx \\ &= \frac{x^2}{2} - 2x + \log|x| + C. \end{aligned}$$
   $x = x^1$ uses the power rule with $n = 1$⁠. The constant $2$ integrates to $2x$⁠. The term $\frac1x = x^{-1}$ is the single case $n = -1$ excluded from the power rule (it would divide by $n + 1 = 0$⁠); its integral is $\log|x|$⁠.
   Standard formulae, items (i) and (xii), Section 7.2; Property (V), Section 7.2.1. Basics: logarithms and 1/x

4. Simplify for the domain

   On our domain $x > 0$ we have $|x| = x$⁠, so the answer can also be written

   $$\frac{x^2}{2} - 2x + \log x + C.$$
   The absolute value in $\log|x|$ is there so that the formula also works for negative $x$⁠. Here $x$ is always positive, so $\log|x|$ and $\log x$ are the same function. Both forms are correct.
   $|x| = x$ for $x > 0$⁠. Basics: logarithms and 1/x

5. Check: differentiate the answer back

   $$\begin{aligned} &\frac{d}{dx}\left(\frac{x^2}{2} - 2x + \log x + C\right) \\ &= x - 2 + \frac1x \\ &= \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2,\ x > 0. \;\checkmark \end{aligned}$$
   The derivative returns the expanded integrand, which Step 2 showed equals the original square for every $x > 0$⁠.
   Property (I), Section 7.2.1; $\frac{d}{dx}\log x = \frac1x$ (Theorem 5, Section 5.4, Chapter 5). Basics: integration undoes differentiation

Every fact used, and where it comes from