# Integration by Parts: When can you not use the Table Method. Why?

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).
• The short answer to your question is "when the integral isn't on the table". Sep 1, 2014 at 18:53
• @Nameless, what do you mean by when it isn't on the table?
– Matt
Sep 1, 2014 at 18:57
• @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).
– Matt
Sep 1, 2014 at 20:24
• 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! Sep 2, 2014 at 5:40
• @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.
– Matt
Sep 2, 2014 at 17:09

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.
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.