I'm learning about integration by parts, primarily from Stewart's text (7th edition). In a supplemental book I have it brings up something called the Table Method. I really find this method appealing because it looks easier and quicker on many problems. But I find it doesn't seem to work at all on some problems (maybe I'm wrong?).

For example, consider: $\int ln(x)\space dx $ (I realize this is an easy one, but I wanted to try the Table Method on it).

Using the table method I get:

\begin{array}{|c|c|c|} \hline u& dv & \pm \\ \hline ln(x)& dx & +1\\ \hline 1/x & x & -1 \\ \hline -1/x^2 & x^2/2 & +1 \\ \hline & & -1 \end{array}

It's apparent that u will never differentiate to zero. I do however see the start of the right answer in this problem using the first diagonal: ($x \ln x$). It seems that in other problems of this type I can do something special with the last full row when I can't get to zero. I can simply integrate the product of the first two cells of that row.

So this would give me first diagonal, plus second diagnal, plus integral product: $x \ln x - {1 \over 2}x + \int-{1\over2}\space dx$

This then works out to $x \ln x - x + C$, which is the correct answer.

I started writing this question thinking that $\int ln(x)\space dx $ can't be solved with the Table Method, but in the process of composing this question it appears it can be in a similar way $\int e^x \cos x \space dx$ can be solved (at least with regards the handling of the last full row going across as an integral instead of diagonal as a non-integral). It took me a while to type all this so I thought I'd still go ahead and post it.

I have some question though:

  1. Please verify if I'm correct.
  2. When you realize the u column will never hit zero, when do you stop? From what I can read it seems you stop once you try differentiating twice. Is this always the case?
  3. I would also like to know if there are problems for which this method cannot work. For example when I try $\int \arctan x \space dx$ I get a mess that doesn't appear to work (but maybe it does and I just don't see how).
  • 1
    $\begingroup$ The short answer to your question is "when the integral isn't on the table". $\endgroup$
    – IAmNoOne
    Sep 1, 2014 at 18:53
  • $\begingroup$ @Nameless, what do you mean by when it isn't on the table? $\endgroup$
    – Matt
    Sep 1, 2014 at 18:57
  • $\begingroup$ @Jean-ClaudeArbaut. Are we confusing looking up antiderivatives from a table of know antiderivative formulas with what I mean by the "table method" (which is something like found here: nebula2.deanza.edu:16080/~lo/2012Spring/1bpartstable.pdf). $\endgroup$
    – Matt
    Sep 1, 2014 at 20:24
  • $\begingroup$ Oh, yes, I didn't know this method at all. If I understand correctly, it's just another way to do integration by parts, shorter when you know there will be several in a row (for example, when there is a $x^n$ in factor of some $f(x)$ and you can differentiate these $x^n$ and integrate the $f(x)$). Looks to me you can always use it, but I'll have to play a bit with it since I just discovered how it works ;-) Sorry for the confusion! $\endgroup$ Sep 2, 2014 at 5:40
  • $\begingroup$ @Jean-ClaudeArbaut, No problem. Glad I was able to introduce you to another way to solve a problem. If you could verify it working (or not working) with $\int \arctan x \space dx$, I'd love to see the solution. $\endgroup$
    – Matt
    Sep 2, 2014 at 17:09

2 Answers 2


You don't need to hit zero to stop, instead, you just stop when you know how to integrate the product of the two cells in the last row, i.e., it is simple enough. Then you can use the Tabular Method as the last step of it is to plus/minus the integration of the product.

For example, in your case, $\int(1/x * x)dx$ is time to stop, and $\int (e^x *(-\cos x))dx$ is good too because it is the negative of what we want to calculate and then we just need to solve a simple equation.

See more about the method: https://en.wikipedia.org/wiki/Integration_by_parts#Tabular_integration_by_parts


I don't know about the "table method", but shall use the opportunity to publicize a notational trick I learned from Peter Henrici, that avoids introducing names $u$, $v$, or similar, for the functions on which we "operate".

Given an integrand which is the product of two expressions $f(x)$ and $g(x)$ (maybe $f(x)\equiv1$) decide which of the two you want to integrate in the process, and write an uparrow $\uparrow$ underneath, and which of the two you want to differentiate, and write a downarrow $\downarrow$ underneath, like so: $$\int\lower6pt\hbox{$\matrix{f(x)\cr\uparrow\cr}$}\lower6pt\hbox{$\matrix{g(x)\cr\downarrow\cr}$}\>dx=F(x)g(x)-\int F(x)g'(x)\>dx\ ,$$ $$\int_a^b\lower6pt\hbox{$\matrix{f(x)\cr\uparrow\cr}$}\lower6pt\hbox{$\matrix{g(x)\cr\downarrow\cr}$}\>dx=F(x)g(x)\biggr|_a^b-\int_a^b F(x)g'(x)\>dx\ ,$$ where $F$ is a primitive of $f$ in the $x$-interval in question.


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