The equation $$ \int e^{x^2} \mathrm{d}x = x e^{x^2} - \int x \cdot 2x e^{x^2} \mathrm{d}x $$ follows by integration by parts.
Consider $ u(x) = e^{x^2} $ and $ v'(x) = 1 $.
Then $$ \int e^{x^2} \mathrm{d}x = \int u(x)v'(x) \mathrm{d}x = u(x)v(x) - \int u'(x)v(x) \mathrm{d}x = x e^{x^2} - \int x \cdot 2x e^{x^2} \mathrm{d}x $$
Now regarding to the integral $$ \int \log( \log x ) - \frac{\mathrm{d}x}{\log x} $$
Firstly notice that what the author wrote is
This is nothing else than using the product rule backwards, however I often find it easier to look at this way.
$$ \int \log( \log x ) - \frac{\mathrm{d}x}{\log x} $$
Meaning that the author is commenting about the preceding text and not about the following integral. In contrast to what you understood
However I find it easier to look at it this way:
Where the colon (that is not present in the original text) suggest that what he supposedly finds easier to look is at which follows.
Which in turn would be $$ \int \log( \log x ) - \frac{\mathrm{d}x}{\log x} $$
As you correctly observed, $\mathrm{d}x$ is not well placed. It is not clear by the context or defined how we should evaluate such expression.
Even if we interpret it as $$ \int \left( \log( \log x ) - \frac{1}{\log x} \right) \mathrm{d}x $$ its evaluation cannot be expressed in terms of elementar, integral free, functions.
Hence I guess this integral besides mistyped is there by mistake and should be ignored.