integration by parts

A technique for solving some indefinite integrals. Based on nothing more than the derivative law for products:

(1)  (u*v)' = u'*v + u*v',

it can solve a great many integrations.

To derive the formula, rewrite (1) as

(2)  u'*v = (u*v)' - u*v'.

Integrating both sides, we see that

(3)  integral u'*v dx = u*v - integral u*v' dx

where we hope u*v' is easier to integrate than u'*v.

A more concise way of writing (3) is to use the formal notation du=u'dx, dv=v'dx, and use that to re-write (3) as

(4)  integral v du = u*v - integral u dv

Integration by Parts
after Henry Reed's
Naming of Parts

To-day's integration by parts. Yesterday,
We did frustums and donuts. And to-morrow morning,
We shall learn partial fractions. But to-day,
To-day's integration by parts. Squirrels
Scamper around by the building,

And to-day's integration by parts.

This is the indefinite integral. And this
Is the definite integral, whose use you will see,
When you have to do areas. And this is the loop integral,
Which in your case, you do not get. The birds
Swoop and flutter, rejoicing in good food and freedom,

Which in your case, you do not get.

This is the homework, which is always completed
By the next day's class. And please do not let me
See anyone writing in pen. You can do it quite neatly
If you have a pencil or two. The oak tree
which stood out the window is gone, made into

A pencil or two.

And this is a syllabus. The purpose of this
Is to give you more problems, as you see. We can make it
Change with a breath and a nudge: we call this
Adapting to demands. With a breath and a nudge,
The wind makes the snow fall and glisten:

They call it adapting to demands.

They call it adapting to demands: it is perfectly easy
If you have an hour or two: like the syllabus,
And the schedule, and the definite integral, and the homework,
Which in your case you did not get; see the squirrels
and the birds getting ready for winter, breathing and nudging each other,

and to-day's integration by parts.

---

--Jurph

Integration by parts is no more than a modification of the product rule for derivatives (you can see a more detailed explanation in ariels’ write-up). The final equation is ∫ u dv = u*v - ∫ v du. As you can see, after using this technique, you will still have an integral to do. The hope is that the resulting integral is simpler than the previous one. The classic example to demonstrate this is ∫x*e^x dx. You want to pick something for u and the rest will be dv. After choosing u and dv, v and du can be found by taking the integral of dv and the derivative of u, respectively.

∫x*e^x dx       u=x      v=e^x                     du=dx   dv=e^x dx Now simply plug in for u, v, du, and dv. x*e^x - ∫e^x dx The integral of e^x is not only one we are able to do, it is also very easy. x*e^x - e^x + c

It is possible to choose the wrong thing for u and dv, but the idea here is to get an integral that is easier than the first, so if the second integral looks harder, it is probably a good idea to start over and choose something else. For example, what if, in the previous example, u was chosen to be e^x instead of x.

∫x*e^x dx       u=e^x          v=(1/2)*x²                      du=e^x dx   dv=x dx (1/2)x*e^x - ∫(1/2)x²e^x dx

This integral looks worse than the first one, and will continue to look worse if you were to keep doing integration by parts. The wrong u has clearly been picked and the problem should be started over.

The question in integration by parts is always what to pick for u. Pick the wrong thing, and you may never finish without starting over from the beginning. Choose the correct substitution and you could be done within one or two steps. How can one know what to choose? I have seen many a Calculus student (myself included) stare at an integral for hours, wondering which part to substitute for u. Fortunately, there is a lovely little mnemonic device to help the bewildered Calc student past the first, and most scary, step: L. I. A. T. E.

LOGRITHEMS INVERSE TRIGONOMETRY FUNCTIONS ALGEBRAIC TRIGONOMETRY FUNCTIONS EXPONETIALS

When looking at an integral that can be solved by integration of parts, select the logarithm for u. If there aren’t any logarithms in the problem, choose the inverse trig function; if there aren’t any inverse trig functions….etc. Simply continue down the list until you have something to substitute in for u. This mnemonic will work most of the time. How about another example to show the mnemonic really does work.

∫x*ln(x) dx         u=ln(x)      v=(1/2)x²                           du=1/x dx   dv=x dx ln(x)*((1/2)x²) - ∫(1/2)x² * (1/x) dx (1/2)x²*ln(x) - ∫(1/2)x dx (1/2)x²*ln(x) – (1/4)x² + c

The first time I saw this particular problem, I thought I should set u equal to x, since it would make du=dx, thereby making the integral much easier. However, doing this would mean taking the integral of the ln(x), a much more complicated thing, which itself must be done using the integration by parts method. “How?” some may ask. “You must always have two things multiplied together to use integration by parts,” they will protest. The integral of the ln(x) does have two things multiplied together. Just watch.

∫ln(x) dx         u=ln(x)        v=x                         du=1/x dx   dv=dx x*ln(x) - ∫x*(1/x) dx x*ln(x) - ∫1 dx x*ln(x) – x + c

Lots of things can be integrated using integration by parts by itself, or in combination with other techniques of symbolic integration. If you’re bored, and want to try this out yourself (or just enjoy torture), here are some more integrals that can be done just by using integration by parts.

∫x*sin(x) dx ∫(ln(x)) / (x³) dx ∫x³*e^-x2 dx

---

Source: My college Calc II class The last trial integral is a little tricky. You must use some creative intuitions to pick the correct values for u and dv.

My Calc 1 teacher had a very nice, clean way of doing repeated integration by parts. By "nice, clean way," I mean that it takes less writing and more doing. As trivia, this "table method" was briefly seen in the movie Stand and Deliver
First off, do the LIATE thing from Cimorene's writeup, to determine what you use for u. This technique seems to be useful when you have an xⁿ in front of some other function whose derivative or antiderivative never "goes to 0" (You can keep differentiating xⁿ and the derivative will go to 0, but if you keep differentiating e^x, the derivative never goes to 0). Here's how the table works:
Three columns, one labeled "u," the second "dv," and the third "sign" or "+/-" or whatever. Or unlabeled. The third column is used for the sign.

    u    |   dv    | sign
--------------------------
         |         |
         |         |
         |         |
         |         |
         |         |
         |         |
         |         |

The u column is where you list all your u-substitutions in terms of x, and the dv is where you list all your dv substitutions in terms of x. Here's how this works:
The nth row in the table will contain the (n - 1)th derivative of the initial u in the "u" column, and the (n - 1)th ANTIderivative of the initial dv in the "dv" column (e.g. the first row will have your initial u and your initial dv. The next row will be the derivative of u and the antiderivative of dv. Next row has the second derivative of u, and second antiderivative of dv. and so on). The u-column should have values until it becomes 0. At this point, you should have the same number of values in the "sign" and the "dv" column.
The sign column works like this: Starting from the second row, put a "+" sign. Next row is a "-". Next is a "+". And so on. It alternates like that.
To use this table to find the integral, you take the value in the first row of the "u" column, multiply it by the SECOND value in "dv", and the SECOND value in "sign". That's the first term of the integral. The next term will be the SECOND row of the "u" column, times the THIRD row of the dv column times the THIRD row of the "sign" column. Ad infinitum, or at least until you're done.

Okay, so that might be confusing. Let's work out a problem:

∫ x³ cos(3x) dx

Since we know the derivative of x³ will eventually go to 0, we say our initial u is x³ and our initial dv is cos(3x). Here's what we have so far:

    u    |     dv      | sign
------------------------------
   x3    |   cos(3x)   |
         |             |

Nothing in the sign column yet. Here's the next row:

    u    |   dv        | sign
------------------------------
   x3    | cos(3x)     |
  3x2    | sin(3x)/3   |  +

Because the antiderivative of cos(3x) = sin(3x)/3, and the derivative of x³ = 3x².
I'll go ahead and fill out the rest of the table:

    u    |   dv        | sign
------------------------------
   x3    | cos(3x)     |
  3x2    | sin(3x)/3   |  +
  6x     | -cos(3x)/9  |  -
  6      | -sin(3x)/27 |  +
  0      | cos(3x)/81  |  -

Make sense? Good? Okay.
(NOTE: If anyone knows how I could improve the ASCII here, help would be appreciated. I'd like to be able to show the lines in the table which connect the products)
The first term is x³ sin(3x)/3 (First row of "u" times second row of "dv" times second row of "sign"). The second term will be -1 ⋅ (3x² ⋅ -cos(3x)/9) = x² cos(3x)/3 (second row of "u" times third row of "dv" times third row of "sign"). Our third term is 6x ⋅ -sin(3x)/27 = -2 x sin(3x) / 9. And so on. Thus:

∫ x³ cos(3x) dx = x³ sin(3x) / 3 - (3x² ⋅ -cos(3x) / 9) + (6x ⋅ -sin(3x) / 27) - (6 ⋅ cos(3x) / 81)
= x³ sin(3x) / 3 + x² cos(3x) / 3 - 2x sin(3x) / 9 - 2 cos(3x) / 27

Voila. Makes integrating things like this less of a hassle.