Example 18: $\int\log x\,dx$

Basics you need

Solution

1. Write $\log x$ as a product and choose the functions

   Write $\log x = (\log x)\cdot 1$ and take

   $$f(x) = \log x \ \ (\text{first}), \qquad g(x) = 1 \ \ (\text{second}).$$

   Then $f'(x) = \frac1x$ and $\int g(x)\,dx = \int 1\,dx = x$⁠.

   We cannot take $\log x$ as the second function, because that would require $\int\log x\,dx$⁠, which is what we are trying to find. As the first function it only needs to be differentiated, and its derivative $\frac1x$ is simple. This matches the book's Remark (iii): a logarithmic function is taken as the first function.
   $\frac{d}{dx}\log x = \frac1x$⁠, $x > 0$⁠. Basics: logarithms and 1/x

2. Apply integration by parts

   $$\begin{aligned} &\int(\log x\cdot 1)\,dx \\ &= \log x\int 1\,dx - \int\Big[\frac{d}{dx}(\log x)\int 1\,dx\Big]dx \\ &= (\log x)\cdot x - \int\frac1x\cdot x\,dx \\ &= x\log x - \int 1\,dx \\ &= x\log x - x + C. \end{aligned}$$
   The product $\frac1x\cdot x$ is just 1 (for $x > 0$⁠, so $x \ne 0$⁠), and $\int 1\,dx = x$⁠. The new integral is trivial, which is the sign of a good choice.
   $\frac1x\cdot x = 1$ for $x \ne 0$⁠; $\int dx = x + C$⁠. Lesson: integration by parts

3. Check: differentiate the answer

   By the product rule, $\frac{d}{dx}(x\log x) = 1\cdot\log x + x\cdot\frac1x$⁠, which is $\log x + 1$⁠. So

   $$\frac{d}{dx}(x\log x - x) = \log x + 1 - 1 = \log x.$$
   The derivative is the integrand for every $x > 0$⁠, so the answer is right. (The answer can also be written $x(\log x - 1) + C$⁠.)
   Product rule (Chapter 5). Basics: integration undoes differentiation

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