SOLUTION: Hi, I was wondering if someone could help me with this question. Evaluate the following indefinite integral using integration by parts: (ln^(2)(x)) sqrt(x) dx Thanks, Jud

Algebra ->  Equations -> SOLUTION: Hi, I was wondering if someone could help me with this question. Evaluate the following indefinite integral using integration by parts: (ln^(2)(x)) sqrt(x) dx Thanks, Jud      Log On


   

Question 226356: Hi, I was wondering if someone could help me with this question.
Evaluate the following indefinite integral using integration by parts:
(ln^(2)(x)) sqrt(x) dx
Thanks, Judy
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The basis for integration by parts is the product rule of differentiation:
d(uv) = u*dv + v*du
This can be rewritten as:
u*dv = d(uv) - v*du
If we integrate this we get:
int%28u%2C+dv%29+=+int%28d%28uv%29%2C+dx%29+-+int%28v%2Cdu%29
int%28u%2C+dv%29 = uv - int%28v%2C+du%29
which is the formula for integration by parts. In order to use this formula we are looking to express an integral as a product of some function we'll call "u" and the derivative of some other function we'll call "v".
In looking at int%28%28ln%28x%29%29%5E2%2Asqrt%28x%29%2C+dx%29 we want to figure out our "u" and "dv". Often the key is to pick "dv" wisely. Since the other side of the formula has 2 v's, we want "dv" to be something easily integrated. Also, since the formula has int%28v%2Adu%29 on the right, we want v*du to an integral that is easier to find than the one we started with.
Since powers of x are easily integrable and since sqrt%28x%29+=+x%5E%281%2F2%29 we will try dv+=+x%5E%281%2F2%29dx. This makes v+=+%282%2F3%29x%5E%283%2F2%29 and u+=+%28ln%28x%29%29%5E2 and du+=+2%2Aln%28x%29%2A%281%2Fx%29. Substituting these into our formula:
=
The integral we still have is not easy. But it is easier than the one we started with. In fact it is very similar to the one we started with. Except it has ln to the first power (instead of squared). Integration by parts can be used to "reduce" the complexity of an integral and it can be used repeatedly to reduce the integral until it is one that can be integrated more easily by another method of integration.
We can use integration by parts on int%28x%5E%281%2F2%29%2Aln%28x%29%2C+dx%29. Let's use dv+=+x%5E%281%2F2%29dx and v+=+%282%2F3%29x%5E%283%2F2%29 again. This time u+=+ln%28x%29 and du+=+%281%2Fx%29dx:

(Note how, after this second integration by parts, that we ended up with an "easy" integral:
Simplifying we get:

And if you differentiate this you get, believe it or not,: %28ln%28x%29%29%5E2%2Asqrt%28x%29!