TITLE: Is there a "more easier" way to integrate by parts?
QUESTION [6 upvotes]: I found a photo from Facebook (now removed) with the following content:

When integrating by parts, students often struggle with how to break up the original integrand into $u$ and $dv$. $\color{blue}{\rm LIATE}$ is an acronym that is often used to determine which part of the integrand should become $u$. Here's how it works: let $u$ be the function from the original integrand that shows up on the list below.

Logarithmic functions e.g. $(\ln x)$
Inverse trigonometric functions e.g. $(\tan^{-1}x)$
Algebraic functions e.g. $(x^{3} + x - 2)$
Trigonometric functions e.g. $(\cos x)$
Exponential functions e.g. $(e^{x})$

In general, we want to let $u$ be a function whose derivative $du$ is both relatively simple and compatible with $v$. Logarithmic and inverse trigonometric functions appear first in the list because their derivatives are algebraic, so if $v$ is algebraic, $v\,du$ is algebraic and an integration with "weird" functions is transformed into one that is completely algebraic. Note that the LIATE approach does not always work, but in many cases it can be helpful.

I tried to use this approach on a relatively simple integral, which is
$$\int (\ln x)^{2}\,dx.$$
I am quite confused whether $u$ should be $\ln x$ or $(\ln x)^{2}$. Is there a more refined way to integrate by parts?

Edit to avoid confusion: I found the antiderivative of $(\ln x)^{2}$, but I am asking for an "easier" way.

REPLY [4 votes]: Here's an old tip for IBP that's useful surprisingly often: when in doubt, take $v=x$ so $u$ is just the existing integrand, and the integration of it becomes closely relation to its easier differentiation. This special case is $\int udx=ux-\int u^\prime xdx$, and for $u=\ln^2x$ the final, subtracted integral is $\int2\ln x dx$, which you can also evaluate by IBP, again with $v=x$ (but of course this time $u$ is $\ln x$ instead of $\ln^2x$).