# Integration by parts

1. Jan 22, 2008

### lgen0290

The teacher kinda skipped over this and expects us to know how to do it. Being I can't understand what he is saying, I'm kinda lost, so I missed how to do the whole u, du thing.

1. The problem statement, all variables and given/known data
Solve $$\int{ 8x^{3}\ln\!\left(x\right) \, dx}$$ using Integration by Parts.

2. Relevant equations

3. The attempt at a solution
Would u by $$8x^{3}$$?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 22, 2008

### G01

Why do you pick u= 8x^3? What's you reasoning behind that choice?

3. Jan 22, 2008

### lgen0290

I am not sure. I picked it as it seemed to be easiest to derive the derivative from.

4. Jan 22, 2008

### G01

It is easy to differentiate, but the problem is that ln(x) is kinda hard to integrate, so I wouldn't let that be dv.

Try letting u = ln(x) and see if you can get to an answer.

5. Jan 22, 2008

### lgen0290

The only problem is I don't do where to go once I get u, du, v and dv. Is it like substitution or?

6. Jan 22, 2008

### rocomath

$$I=uv-\int vdu$$

Last edited: Jan 22, 2008
7. Jan 22, 2008

### G01

This formula is incorrect. It should read:

$$I=uv-\int vdu$$

Under the integral, you should have vdu, not udv.

Anyway, once you have u,v,du,dv, you can use the above formula. It will be equivalent to the integral given to you.

8. Jan 22, 2008

### rocomath

9. Jan 23, 2008

### fermio

$$\int u dv=uv-\int vdu$$

$$\int 8x^3\ln x dx=8(\frac{x^4}{4}\ln x-\int \frac{x^4}{4}\cdot\frac{1}{x}dx)=x^4(2\ln x-\frac{1}{2})+C[\tex] [tex] u=\ln x$$
$$v'=x^3$$
$$u'=\frac{1}{x}$$
$$v=\frac{x^4}{4}$$

Last edited: Jan 23, 2008
10. Jan 23, 2008

### HallsofIvy

Staff Emeritus
Fermio, is there any point in just doing the problem yourself rather than letting Igen0290 try it after the suggestions?

11. Jan 23, 2008

### G01

Yes, fermio, what was the point of doing the problem for someone else? It is very easy to just give an answer, and it does not provide actual help to the OP. Yes, they get the answer, but do they understand? Do they actually learn how to solve problems of that type independently? Also, when you post a complete solution like that, you negate the help we were trying to provide to the student. Please think before you post a complete solution next time.

12. Jan 23, 2008

### lgen0290

Can someone explain it please? How do you determine which is u and which is dv?

13. Jan 23, 2008

### awvvu

We were taught to use LIPET -- logarithms, inverse trig, polynomial, exponential, trig. Take u to be something on the left and dv something on the right.

It's on wikipedia's page for integreation by parts too, though a bit different.

http://en.wikipedia.org/wiki/Integration_by_parts#The_ILATE_rule

The exact order doesn't really matter; just know to pick a logarithmic or inverse trig function for u rather than a polynomial.