What is the best approach for Integration by parts?

Click For Summary

Homework Help Overview

The discussion revolves around the technique of integration by parts, specifically focusing on how to choose the appropriate functions for \( u \) and \( dv \) in various integral problems involving logarithmic and exponential functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants share their attempts at applying integration by parts to integrals such as \( \int (ln(x))^2 \) and \( \int x^2 (\ln x)^2 dx \), discussing their choices for \( u \) and \( dv \). Some express confusion about how to effectively split the integral and the reasoning behind their choices.

Discussion Status

Several participants have provided insights and shared their thought processes, with some expressing uncertainty about their selections for \( u \) and \( dv \). There is an ongoing exploration of different integrals, and while some participants feel they have made progress, others continue to seek clarification on their approaches.

Contextual Notes

Participants mention the challenges of integrating functions where the power increases after applying integration by parts, leading to recursive situations. There is also a reference to a mnemonic for choosing \( u \) based on function types, indicating a shared understanding of common strategies in integration by parts.

Tom McCurdy
Messages
1,021
Reaction score
1
I am usually alright once I figure out how to split up the integral into
u:
du:
v:
dv:
so i can simply do
[tex]uv-\int v*du[/tex] but I keep messing up on there I will post some examples if I can find them and if someone could help me that would be great

[tex]\int (ln(x))^2[/tex]
 
Physics news on Phys.org
Here is what I am trying

[tex]\int (ln(x))^2[/tex]

u: ln x
du: 1/x
v: x ln(x)-x
dv: ln x

[tex](x ln(x)-x)ln(x)-\int {ln(x)}-1[/tex]

[tex](x ln(x)-x)ln(x)-(x ln(x) -x) + x[/tex]
 
Last edited:
I am also trying to figure out u and dv for [tex]\int (x ln x)^2[/tex] but i figured it would follow in line once i figured out the previous problem
 
Last edited:
Hey tom,

For

[tex]\int x^2 (\ln x)^2 dx[/tex]

Use [itex]u = (\ln x)^2[/itex]
 
Last edited:
Tom McCurdy said:
Here is what I am trying

[tex]\int (ln(x))^2[/tex]

u: ln x
du: 1/x
v: x ln(x)-x
dv: ln x

[tex](x ln(x)-x)ln(x)-\int {ln(x)}-1[/tex]

[tex](x ln(x)-x)ln(x)-(x ln(x) -x) + x[/tex]


You have the right answer, what's the problem?
 
quasar987 said:
You have the right answer, what's the problem?
I figured it out and kept editing it... thanks though... I am still not sure how I would know to split up the U and dv in the way that I did
 
ok I got it down to
[tex](x^2 ln(x))(x ln(x) - x) - [ \frac{x^3(3ln(x)-1)}{27}-\frac{x^3}{27} ][/tex]
 
out of curisoity is there any rules on how to do [tex]\int \frac{x^2 ln(x)}{27}[/tex] because it was needed for the previous problem and I just used my calculator to come out with the first fraction after the first bracket.
 
That's an integration by parts

If we rearrange it:

[tex]\frac{1}{27} \int x^2 \ln x dx[/tex]
 
  • #10
Tom McCurdy said:
I am still not sure how I would know to split up the U and dv in the way that I did

Unfortunately, I don't think there's any other way than trial an error.
 
  • #11
Quasar is right, but you will learn to notice which to pick as you make plenty integration by parts.
 
  • #12
thx... what would I do to start [tex]\int tan^{-1}[/tex]
 
Last edited:
  • #13
oh wait the derivative of [tex]tan^{-1} = \frac{1}{(x^2+1)}[/tex] So I set tan^-1 as u and 1 as dv... let me see what i get
 
  • #14
Assuming your mean arctg

I recommend [itex]u = arctg[/itex]
 
  • #15
I was doing the [tex]\int x sin^{-1}[/tex] and was wondering once you got to
[tex]x sin^{-1} - \int \frac{x}{\sqrt{1-x^2}}[/tex] how you would integrate [tex]\int \frac{x}{\sqrt{1-x^2}}[/tex] I know the answer is [tex]-\sqrt{1-x^2}[/tex] but how do you get that
 
  • #16
Alright
I am doing the [tex]\int x^2 e^{5x}[/tex] and I got to here
assuming u:x^2 and dv: e^(5x)
[tex]\frac{x^2*e^{5x}}{5} - \frac{2}{5} \int 2x e^{52}[/tex]

what would I do now?
 
  • #17
Assuming you mean 5x instead of 52.

Integrate by part that integral :smile:
 
  • #18
oh ****... this is going to take forever... I am on problem 18 out of 64 and I realized that the problems ahead will require 5 integration by parts
 
  • #19
lol this is what i mean [tex]\int x^5*e^{2x^3}[/tex]
 
  • #20
woah this was werid...

i was doing [tex]\int x^3*e^{x^2}[/tex] and I got down to

[tex]2x^4 e^{x^2}-2 \int x^3*e^{x^2}[/tex] which is the original problem... how would I go about solving this...
 
  • #21
Since

[tex]\int x^3*e^{x^2} = 2x^4 e^{x^2}-2 \int x^3*e^{x^2}[/tex] can I just say
[tex]3 \int x^3*e^{x^2} = 2x^4 e^{x^2}[/tex]
[tex]\int x^3*e^{x^2} = \frac{2x^4 e^{x^2}}{3}[/tex] ?
 
  • #22
Yes you can.
 
  • #23
It's True! Lazy People Ride The Elevator!

I was always taught to pick [tex]u[/tex] by the following priority,…

Inverse Trig
Logarithms
Polynomials
Rational
Trigonometric
Exponential'

The order can be remembered by the saying at the top of this post. :biggrin:

In the case that you were working with here you should let [tex]u[/tex] equal the natural log function and then let [tex]dv[/tex] equal what's left.

This system may not be fail-proof, but it seems to work most of the time. Actually it's never failed me yet.
 
  • #24
thx... I am having trouble with [tex]\int x^5*e^{2x^3}[/tex]

what should be U and what should be dv...

I choose u:x^5 and dv: e^(2x^3) but it got out of hand quickly
 
  • #25
oh shoot... I see a mistake I made... I had it as
[tex]\int x^5*e^{{(2x)}^3}[/tex]
instead of
[tex]\int x^5*e^{2x^3}[/tex]
 
  • #26
nope still have a problem when I use u: x^5 and dv: e^(2x^3)
the power increases and after first one you get

[tex]6x^7 e^{2x^3} - 30 \int x^2 e^{2x^3}[/tex]... what should i do
 
  • #27
Keep on integrating, and pray you didn't make a mistake

Also a note: I've noticed you don't like putting dx in your integrals, do it on your work, some teachers (specially mathematicians) take points off if you don't.
 
  • #28
Tom McCurdy said:
nope still have a problem when I use u: x^5 and dv: e^(2x^3)
the power increases and after first one you get

[tex]6x^7 e^{2x^3} - 30 \int x^2 e^{2x^3}[/tex]... what should i do

sorry i mistyped last time
the reason i am nervous that i did something wrong is that the power inside the integral increased instead of decreased
it went from 5 to 6
correct version
[tex]6x^7 e^{2x^3} - 30 \int x^6e^{2x^3}[/tex]...
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 69 ·
3
Replies
69
Views
7K
Replies
1
Views
3K
  • · Replies 27 ·
Replies
27
Views
4K