Exercise 7.2 · Q2: $\int \frac{(\log x)^2}{x}\,dx$

Basics you need

Solution

1. Split the integrand to expose $\frac{1}{x}$

   $$\frac{(\log x)^2}{x} = (\log x)^2\cdot\frac{1}{x}.$$
   Written this way the integrand is "a function of $\log x$" times "the derivative of $\log x$". That is the exact pattern that substitution handles: the inner expression is $\log x$⁠, and its derivative $\frac1x$ is present as a factor.
   $\frac{d}{dx}\log x = \frac1x$⁠, $x > 0$⁠. Basics: logarithms and 1/x

2. Substitute $t = \log x$

   $$\begin{aligned} t &= \log x, \\ \frac{dt}{dx} &= \frac{1}{x}, \\ dt &= \frac{dx}{x}. \end{aligned}$$

   Therefore

   $$\int (\log x)^2\cdot\frac{dx}{x} = \int t^2\,dt.$$
   The factor $(\log x)^2$ becomes $t^2$⁠, and the leftover $\frac{dx}{x}$ is exactly $dt$⁠. Every $x$ has been replaced, so the substitution is complete. As $x$ runs over $(0, \infty)$⁠, $t = \log x$ runs over all real numbers, and $t^2$ is defined for all of them.
   Substitution, Section 7.3.1. Lesson: integration by substitution

3. Integrate with the power formula

   $$\int t^2\,dt = \frac{t^{2+1}}{2+1} + C = \frac{t^3}{3} + C.$$
   The power formula with $n = 2$ (allowed, since $n \ne -1$⁠): raise the power by one and divide by the new power.
   Standard formula (i), Section 7.2. Basics: the power rule

4. Return to $x$

   $$\frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C, \qquad x > 0.$$
   Replace $t$ by $\log x$⁠. The answer must be stated in the original variable, on the original domain $x > 0$⁠.

5. Check: differentiate the answer

   Chain rule with inner $u = \log x$ and outer $\frac{u^3}{3}$⁠:

   $$\begin{aligned} \frac{d}{dx}\left[\frac{(\log x)^3}{3}\right] &= \frac{3(\log x)^2}{3}\cdot\frac{d}{dx}(\log x) \\ &= (\log x)^2\cdot\frac{1}{x} \\ &= \frac{(\log x)^2}{x}. \end{aligned}$$

   We get back the integrand, so the answer is correct for every $x > 0$⁠.

   Differentiation undoes integration (Property (I), Section 7.2.1), so getting the integrand back confirms the antiderivative.
   Power rule $\frac{d}{du}u^3 = 3u^2$ and the chain rule (Chapter 5, Theorem 4). Basics: integration undoes differentiation

6. Compare with the textbook's form

   The answer key prints $\frac13(\log|x|)^3 + C$⁠. On our domain $x > 0$ we have $|x| = x$⁠, so

   $$\frac{1}{3}\left(\log|x|\right)^3 = \frac{(\log x)^3}{3}.$$

   The two answers are identical.

   The key's absolute value does no harm: it is the form that would also cover negative $x$ if the integrand were read as $\frac{(\log|x|)^2}{x}$⁠. For the function as printed, $\log x$ needs $x > 0$⁠, and there both forms agree.

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