Integration by parts VS. Formula in text

In summary: So you are correct. You just needed to factorise the 2 from the term you were uncomfortable with.In summary, the conversation involves a person trying to find the correct integration for the formula \int\ln u du=u\ln u-u+C using the integration by parts method. They encounter a sign error and difficulty in factoring, but ultimately arrive at the correct answer of \frac{1}{2}[(2x-3)\ln(2x-3)-(2x-3)]+C by factoring out the constant term and correcting the sign error.
  • #1
Saladsamurai
3,020
7
I am either making the same mistake repeatedly, or I can't factor!

I did this [tex]\int\ln(2x-3)dx[/tex] first by parts and then using the formula [tex]\int\ln u du=u\ln u-u+C[/tex]

By parts I got:
[tex]u=\ln(2x-3)[/tex] dv=dx

so [tex]=x\ln(2x-3)-\int\frac{2x}{2x-3}dx[/tex] and by long division:

[tex]=x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]

[tex]=x\ln(2x-3)-x-\frac{3}{2}\ln(2x+3)+C[/tex]



but I can't get that to match the formula result of

[tex](2x-3)\ln(2x-3)-(2x+3)+C[/tex]

Is the by parts correct?

Casey
 
Physics news on Phys.org
  • #2
I am getting closer, but I must have screwed up a sign or something in the Int by parts...but I can't see where ...
 
  • #3
that's what i got, let me take the derivative and to check.
 
Last edited:
  • #4
Okay, so I have fixed a sign error and I get from parts:

[tex]x\ln(2x-3)-x-\frac{3}{2}\ln(2x-3)[/tex] and then multiplying thru

by 2 I get [tex]2x\ln(2x-3)-2x-3\ln(2x-3)[/tex] which is extremely close...but I can't figure out how to factor it properly...

Almost there :)

I just can't figure out to do with that -2x in the "middle" its a term not a factor...
 
  • #5
For one,

[tex]x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]

should have a minus sign within the integral.

In the following step, you've multiplied the - outside the integral(that's supposed to be there :wink:) with that +.
 
  • #6
it checks out after taking the derivative.
 
  • #7
neutrino said:
For one,

[tex]x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]

should have a minus sign within the integral.

In the following step, you've multiplied the - outside the integral(that's supposed to be there :wink:) with that +.

I don't follow. If I do that integral off to the side I get x+(3/2)ln(2x-3)

and then i distibuted the - sign...

EDIT: I see, I just forgot to put it in the 1st post...scroll down to post#4
 
Last edited:
  • #8
rocophysics said:
it checks out after taking the derivative.

What does? from which post#. I can't get it to match the formula.

Casey
 
  • #9
first post, except that the last ln should be ln(2x-3)

i hate the tables, i prefer hand-method :p takes too long and comes with errors because i suck, but is all good.
 
  • #10
My goal though is to get what I got in post #4 to look like the formula from the textbook which looks like this [tex](2x-3)\ln(2x-3)-(2x+3)+C[/tex]

but I have [tex]2x\ln(2x-3)-2x-3\ln(2x-3)+C[/tex]

which is close...how does that factor?

If I factor that don't I just get [tex] (2x-3)\ln(2x-3)-2x+C[/tex] ?
 
Last edited:
  • #11
why can't I factor this:rofl:??!
 
  • #12
lol that's really hard ... quit obsessing!
 
Last edited:
  • #13
rocophysics said:
lol that's really hard ... quit obsessing!

But it is part of the assignment!
 
  • #14
Saladsamurai said:
But it is part of the assignment!
to put it in table form? eek!

i got:

[tex]\ln{(2x-3)}[2x-3]-x[/tex] ...

but idk what to do after that. and how did you get -2x?
 
  • #15
Maybe it's because what you're trying to find is an indefinite integral, and the answers to different methods employed usually differ by a constant.
 
  • #16
This might help:

[tex]-x+C=-\frac{1}{2}(2x)+C=-\frac{1}{2}(2x-3)+C-\frac{3}{2}=-\frac{1}{2}(2x-3)+C_1[/tex]

In which [tex]C_1=C-\frac{3}{2}[/tex] another constant is.
 
  • #17
salad, I can't quote your post, browser issues. But your "book formula" integral is in error.

[tex](2x-3)\ln(2x-3)-(2x-3)+C[/tex]

should read

[tex]\frac{1}{2}[(2x-3)\ln(2x-3)-(2x-3)]+C[/tex]

(I corrected your sign error)

You cannot in general just use a formula for an integral of f(u) with a direct substitution of u=g(x) into the result. This is because what you are actually doing with a proper substitution is

[tex]\int{f(g(x))}dx = \int{\frac{dx}{du}[f(u)]}du = \int{\frac{f(u)}{u'}}du[/tex]

In this case, u'(x) = 2, so you should divide the final expression by two.

I did not check your integration by parts, but the above correct answer for the formula method tallies with it, except for a constant term :

[tex]x\ln{(2x-3)} - x - \frac{3}{2}\ln{(2x-3)} = \frac{1}{2}(2x)\ln{(2x-3)} - \frac{1}{2}(2x) - \frac{1}{2}(3\ln{(2x-3)}) = \frac{1}{2}[(2x-3)\ln{(2x-3) - 2x}] = \frac{1}{2}[(2x-3)\ln{(2x-3) - (2x - 3)}] + C[/tex]

which all works out.
 
Last edited:

1. What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It involves breaking down a complex integral into simpler parts and using a specific formula to solve for the final result.

2. What is the formula used in integration by parts?

The formula used in integration by parts is ∫ u dv = uv - ∫ v du, where u and v are functions of x and dv and du are their respective differentials.

3. How does integration by parts compare to the formula in text?

Integration by parts is a method that uses a specific formula to solve for integrals, while the formula in text refers to the general formula used in calculus to find the integral of a function. Integration by parts is a more specific and step-by-step approach, while the formula in text is a more general concept.

4. When should I use integration by parts versus the formula in text?

You should use integration by parts when the integral you are trying to solve is a product of two functions. The formula in text can be used for any type of integral, but it may be more efficient to use integration by parts for certain types of integrals.

5. What are the benefits of using integration by parts versus the formula in text?

Integration by parts allows for the integration of more complex functions that cannot be solved using the basic formula in text. It also helps to simplify the integral into smaller parts, making it easier to solve. However, the formula in text is a more general concept and can be used for a wider range of integrals.

Similar threads

  • Calculus and Beyond Homework Help
Replies
15
Views
745
Replies
5
Views
936
  • Calculus and Beyond Homework Help
Replies
5
Views
742
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
772
  • Calculus and Beyond Homework Help
Replies
1
Views
950
  • Calculus and Beyond Homework Help
Replies
2
Views
871
  • Calculus and Beyond Homework Help
Replies
4
Views
606
  • Calculus and Beyond Homework Help
Replies
21
Views
1K
Back
Top