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On the article Lesser known integration tricks, I found this:

\begin{align*} J &= \int (1 + 2x^2) e^{x^2} \mathrm{d}x \\ &= \int 2x^2e^{x^2}\mathrm{d}x + \int e^{x^2} \mathrm{d}x \, \\ &= \int 2x^2e^{x^2} \mathrm{d}x + \left[ x e^{x^2} - \int x \cdot 2x e^{x^2} \mathrm{d}x \right] \\ & = x e^{x^2} + \mathcal{C} \end{align*} However I find it easier to look at it this way. $$ \int \log( \log x ) - \frac{\mathrm{d}x}{\log x}$$

I can't see how can the last equation be equal to the first one. Also, isn't d$x$ supposed to multiplying the entire integrand instead of multiplying just $\frac{1}{\log x}$ ?

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    The original answer has "This is nothing else than using the product rule backwards, however I often find it easier to look at this way.". This sentence ends with a full stop and may just refer to the preceding derivation, with the following integral being an unrelated copy-and-paste typo. You've connected (a modified version of) the sentence with the integral following it, possibly changing the meaning. – Frentos Oct 29 '17 at 18:57
  • @Frentos that's right, I unintentionally changed a period for a colon. I fixed that issue – evaristegd Oct 29 '17 at 23:42

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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.

mucciolo
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