Integration by Parts: Solve 8x^{3}\ln\!\left(x\right)

[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.

Homework Statement

Solve \int{ 8x^{3}\ln\!\left(x\right) \, dx} using Integration by Parts.

Homework Equations

The Attempt at a Solution

Would u by 8x^{3}?

 

[G01]

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

 

[lgen0290]

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

 

[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.

 

[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?

 

[rocomath]

I=uv-\int vdu

 

[G01]

I=uv-\int udv

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.

 

[rocomath]

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.

Crap, my bad.

 

[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]<br /> <br /> <br /> u=\ln x <br /> v&amp;#039;=x^3<br /> u&amp;#039;=\frac{1}{x}<br /> v=\frac{x^4}{4}

 

[HallsofIvy]

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

 

[G01]

\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]<br /> <br /> <br /> u=\ln x <br /> v&amp;#039;=x^3<br /> u&amp;#039;=\frac{1}{x}<br /> v=\frac{x^4}{4}

<br /> <br /> <blockquote data-attributes="" data-quote="HallsofIvy" data-source="post: 1581489" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> HallsofIvy said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Fermio, is there any point in just doing the problem yourself rather than letting Igen0290 try it after the suggestions? </div> </div> </blockquote><br /> 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.

 

[lgen0290]

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

 

[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.