Is there a "more easier" way to integrate by parts?

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.

• You want to choose $u$ and $dv$ so that the $dv$ factor is easy to antidifferentiate. If you choose $u = \ln x$, then $dv = \ln x\,dx$, but then $v$ is the antiderivative of $\ln x$, which you might know but is not that simple. So it's better to choose $u = (\ln x)^2$. May 10 '21 at 15:31
• Look up the tableau method. Each row of the table generated represents the integral left to do. In some circumstances this last row is $0$ but it might be a copy of the original row or an integral which can be done. The method takes a lot of the drudge out of the process giving you more time to play with approaches.
– Paul
May 10 '21 at 16:00
• Disclaimer: I am not particularly good at computing antiderivatives. Still, I somewhat have the feeling that the really good people do not rely on these acronyms or similar mnemonic methods. Rather, with practice and with a deeper knowledge of math, they develop their own personal intuition and tricks. I wouldn't spend a lot of time with that acronym. Just my 2 cents. May 10 '21 at 16:09

Yes, here is a general rule that I like to follow. First I will say out some notation: $$I[f]=\int fdx,\,\,\,D[f]=\frac{d}{dx}f$$ and in general we can say: $$\int fg\,dx=fI[g]-D[f]I^2[g]+D^2[f]I^3[g]-...$$ So in general you have this $$+,-$$ pattern and you are differentiating the first function and integrating the second function.

Here is an example: $$\int \underbrace{x^2}_{\color{red}{D}}\underbrace{e^{-ax}}_{\color{blue}{I}}\,dx$$ $$=\color{red}{\left[x^2\right]}\color{blue}{\left[\frac{e^{-ax}}{-a}\right]}-\color{red}{\left[2x\right]}\color{blue}{\left[\frac{e^{-ax}}{a^2}\right]}+\color{red}{\left[2\right]}\color{blue}{\left[\frac{e^{-ax}}{-a^3}\right]}-\color{red}{\left[0\right]}\color{blue}{\left[\frac{e^{-ax}}{a^4}\right]}$$ now we can stop at this point because if we just keep differentiating $$0$$ we still get $$0$$.

You may also wish to think of it as this: $$\int \underbrace{f}_{D}\underbrace{g}_{I}\,dx=D_0I_1-D_1I_2+D_2I_3-...$$

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 $$\ln^2x$$).

The idea here is that we have something we don't know how to integrate, but maybe taking derivatives makes it simple. We don't know an easy integral for $$\ln (x)$$, so we make $$u=(\ln x)^2$$ We also want $$dv$$ something easy to integrate. Again, the only option here for something easy to integrate is $$dx$$

• Sorry, my main question is about choosing the expressions of $u$ and $dv$ that would easily solve the integral, not about the example. May 10 '21 at 15:39
• In general you pick a $u$ that gets simpler when you differentiate it and a $dv$ that doesn't get more complicated
– Alan
May 10 '21 at 15:45
• No, I mean, if there is a known way that is 'better' than the one I presented. May 10 '21 at 15:53

Integration by parts is a bit complicated for the inexperienced.

1)

Therefore writing integration-by-parts rule with the following two formulas is helpful.

$$\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$$ $$\int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)dx$$

So you can look which derivatives and which integrals you can calculate.

Clearly, you can also choose the usual notation with $$u$$ and $$v$$.

But usually, integration by parts is wanted for integrating products.

Integration-by-parts rule can be transformed to the following product rules of integration:

Let $$F(x)=\int f(x)dx+c_1$$, $$G(x)=\int g(x)dx+c_2$$.

$$\int f(x)g(x)dx=f(x)G(x)-\int f'(x)G(x)dx$$

$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx$$

Clearly, you can also choose the usual notation with $$u$$ and $$v$$ - supplemented by $$U$$ and $$V$$.

These product rules and references are written in:

2)

Alcantara 2015 (see the reference below) writes: "The hyperbolic functions can be included from the original LIATE Rule and make it LIATHE Rule."

3)

There is the method called Column Integration or Tabular Integration by Parts (TIBP). Henry Lee's answer describes it.

The newest references I found for Tabular Integration by Parts are:
Alcantara, E. C.: On the Derivation of Some Reduction Formula through Tabular Integration by Parts. Asia Pacific Journal of Multidisciplinary Research 3 (2015) (1) 80-84
Alcantara, E. C.: Integrals of composite functions through tabular integration by parts

The problem of integration by parts is to make the right choice or to make the wrong choice. In Tabular Integration by Parts, you have a function to differentiate successively and a function to integrate successively. So you should make the right choice.

The function terms of the Liouvillian functions, among them the Elementary functions (Differential algebra), are infinitely differentiable. Liouville's theorem and Risch algorithm can help to say which Liouvillian or elementary functions have Liouvillian or elementary antiderivatives.