Answer to Exercise 11.4-3

The problem is to use integration by parts to integrate

x ln(x) dx

Step 1: Choose your parts. There are two ways you could have chosen the parts in this one:

a)
u = ln(x)
and
dv = x dx

or
b)
u = x
and
dv = ln(x) dx

By a fluke, you can get to a correct answer either way on this one. But unlike the first one where choosing a) or b) took you down parallel paths, in this one a) and b) take you down markedly divergent paths, even though they both end up at the same destination. So again I will do it both ways, starting with a), because it is the easier trek. To go right to my working of b), click here.

Step 2: What is v? Remember that you find v by integrating the expression you choose for dv/dx. So do that using the expression given in a) for dv/dx, then click here to see if you got the right expression.

You have dv/dx = 1. So integrating that you get:

v =

x dx =

1 x² + C 2

You remember that in this step you are allowed to drop the C in the above when you stick this in for v into equation 11.4-6a (where you keep 11.4-6a's C). So what you have so far is

1 x² ln(x) + C - 2

1 2

du x² dx = dx

x ln(x) dx

Step 3: What is du/dx? You find that by taking the derivative of u = ln(x), whose derivative you can find on the table. So do that, then plug the result into the above and click here to move on.

Of course you have

   du     1
       =   
   dx     x

and when you put that into the integration by parts you have already developed, you get

1 x² ln(x) + C - 2

1 2

x² dx = x

x ln(x) dx

Of course you can take the cancellation in the integrand that's to the left of the equal and get

1 x² ln(x) + C - 2

1 2

x dx =

x ln(x) dx

Step 4: Do the remaining integral: It's a pretty easy one that we already had to do once in this problem. So I won't make you click for it:

1 x² ln(x) - 2

1 x² + C = 4

x ln(x) dx

Now can you generalize the same method shown above to find the indefinite integral of

xⁿ ln(x) dx

and what is the one value of n for which your generalization won't work? See if you solve that oddball case using simple substitution?

So what if you chose b) when you chose your parts? Then you have as Step 1:

b)
u = x
and
dv = ln(x) dx

Step 2: What is v? You find that by integrating ln(x), which we already did back in the main text. So

   v  =  x ln(x)  -  x  +  C

And you can drop the C when you plug this into equation 11.4-6a.

x² ln(x) - x² + C -

x ln(x) - x

du dx = dx

x ln(x) dx

Step 3: What is du/dx? Just take the derivative of u = x. And that gives you

   du
       =  1
   dx

which simplifies things a little:

x² ln(x) - x² + C -

(x ln(x) - x) dx =

x ln(x) dx

Step 4: Do the remaining integral. Since the integrand of the remaining integral is a difference, you can break it up according to equation 11.2-13b. And (being careful to get the signs right) you get

x² ln(x) - x² + C -

x ln(x) dx +

x dx =

x ln(x) dx

The second integral is one you already know (or can find on the table). Taking that and combining like terms, you get

x² ln(x) -

1 x² + C - 2

x ln(x) dx =

x ln(x) dx

Now you've reached the gotcha where the only integral remaining to be done is identical to the one you started with. But if you add that integral to both sides, then divide the entire equation by 2, you find that you get the same result as doing it the first way.

Like the first method, this approach can also be generalized to integrate

xⁿ ln(x) dx

Try it. And observe that it too fails on the one oddball case for n.

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