Why Do I Struggle to Choose the Correct u and dv in Integration by Parts?

[pillar]

Find
http://img214.imageshack.us/img214/4186/problemm.png

Homework Equations

udv=uv-ƒvdu

The Attempt at a Solution

lndx=dv
(1/X)=v

u=x^2+2
du= 2x^2

I looked in the solutions manual and I don't get, why do I keep picking the wrong u & dv?Can someone please show an example on how to pick the right term for u and dv?

Thanks in advance.

 

[CFDFEAGURU]

From my Calculus text,

Let u = Ln(x) and let dv = x^n

In general always let v be the easier function to integrate.

Thanks
Matt

 

[tiny-tim]

Hi pillar!

(have an integral: ∫ and try using the X² tag just above the Reply box )

Try the obvious … u = 1

 

[union68]

Integration by parts is for PRODUCTS of functions, not COMPOSITIONS. Do you see your error in assigning your u and dv?

The integrand is a composition \ln x and x^2 +2...not a product of the two.

I'll get you started:

On the face of it, we only have ONE function. But if we multiply the integrand by 1, then it's unchanged, right? Well then, we have the product of TWO functions! One is \ln \left(x^2 +2\right) and the other is 1!

Let u = \ln \left(x^2 +2\right) and dv = 1 dx. Then du = 2x / \left(x^2+2\right) and v=x. Then...

 

[tiny-tim]

Hi union68!

Please don't provide full (or almost-full) answers …

just give enough to get the OP going again! ​

(the OP is still offline, so can you edit a bit?)

 

[union68]

pillar mentioned having the solution manual, I highly doubt I gave more info then what was in there...except for the last hint. I'll yank that.

 

[pillar]

pillar mentioned having the solution manual, I highly doubt I gave more info then what was in there...except for the last hint. I'll yank that.

The solution is rather detail (camera not working) and the class hasn't gotten this far yet.

The teacher didn't talk about composition and product laws. Instead what I was able to gather was that you give U to the one easiest to integrate and keep simple to integrate stuff simple.

 

[Elucidus]

This integration requires integration by parts, a polynomial long division, and two different integration substitutions. I am curious why this example was chosen to illustrate integration by parts as it is the least interesting part of the solution.

--Elucidus

 

[pillar]

This integration requires integration by parts, a polynomial long division, and two different integration substitutions. I am curious why this example was chosen to illustrate integration by parts as it is the least interesting part of the solution.

--Elucidus

Maybe to show 1 could be a dv or something.

 

[tiny-tim]

Yes, dv = 1 … carry on from there.

 

[pillar]

Apparently model my book goes by for choosing u is LIATE (Logmetric.Inverse Trig.Algebraical.Trig.Exponential function)

 

[tiny-tim]

Apparently model my book goes by for choosing u is LIATE (Logmetric.Inverse Trig.Algebraical.Trig.Exponential function)

ok … well, add 1 to the list

(it does come up quite often)

 

[union68]

The solution is rather detail (camera not working) and the class hasn't gotten this far yet.

The teacher didn't talk about composition and product laws. Instead what I was able to gather was that you give U to the one easiest to integrate and keep simple to integrate stuff simple.

There are no "composition or product laws" that need to be talked about. The formula for integration by parts is actually derived from the product rule for differentiation, but never mind that. You don't need to know that at this point. Just remember that integration by parts is for integrals of the form


\int f \left(x\right) g \left(x\right) \ dx,

and typically NOT

\int f \left( g \left(x\right) \right) \ dx.

It's usually used where the standard u-substitution from calc II fails. If you wanted to use integration by parts for the second integral (which is like the question you posted!) you CANNOT set u = g \left(x\right) and dv = f \left(x\right). Look at these integrals...

\int x \cos x \ dx

and

\int x \cos x^2 \ dx.

They look similar, but use different techniques. Can you recognize why?

Part of learning all these rules is also learning where and under what conditions each rule can be applied. This is where your error was.