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Integration by parts

  1. Oct 19, 2005 #1
    Hi,

    I'm a bit confused as to what I should assign u and dv in this integration by parts:

    ln(1+x^2)dx

    I remember a general rule called the "LIPATE" rule... which is basically Logarithms, inverse trigs, poly, algebra, trig, then exponentials...

    Now... would I assign u = ln(1+x^2)? and dv = dx?

    That kind of defeats the purpose of integration by parts... doesn't it?

    I'm probably not assigning u and dv properly... so if anybody can lend me a hand, thank you!
     
  2. jcsd
  3. Oct 19, 2005 #2

    Tom Mattson

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    Yes.

    No, it works like a charm. Try it and see.
     
  4. Oct 20, 2005 #3
    Ok... so this is what happens after I work it out

    u=ln(1+x)^2 du=2x/(x^2+1)

    dv=dx v=x

    ln(1+x^2)x - integral [ 2x^2/(1+x^2)

    Using a calc, I found out that the integral works out to be x - arctan(x)

    ... but I don't know how to do it ... any pointers?
     
  5. Oct 20, 2005 #4

    Tom Mattson

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    Your "new" integrand is an improper rational function. They should have beat in into your head during your precalc courses that improper rational functions should be decomposed into a polynomial plus a proper rational function, via long division.

    Try that, and then see if it's not clear what to do.
     
  6. Oct 20, 2005 #5
    I'll need a quick refresher on that... I don't remember exactly how to do that...
    I'll take a look around on the net. In the meantime, I am 100% stuck on this question:

    integral of x^7/(1+x^4)^(3/2)

    I've tried assigning u=x^7 and dv="the bottom part"

    unfortunately... I don't know where I'm headed.

    I'm sorry for all these questions. I have a hard time trying to determine what I should assign to u and dv. Is there a good tutorial online or something that can help me in this area?
     
  7. Oct 21, 2005 #6

    Tom Mattson

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    Don't get so caught up in integration by parts that you forget the other stuff you used to know. This can be done with a simple u-substitution. I'll let you figure out what u is supposed to be.
     
  8. Oct 22, 2005 #7
    Thanks... but this section in the book insists that we use integration by parts in order to solve these questions... (Although using substitution as a secondary method would be acceptable I think).

    I've solved the questions... but I'm now stuck on integration and trigonometric identities...

    Integral of sec(x)^4 / tan(x)^2

    I've tried using sec(x)^2 = tan(x)^2 + 1 and vice versa, but I don't seem to be getting anywhere with that. If someone could possibly lend a few hinters! Thanks!
     
  9. Oct 23, 2005 #8

    VietDao29

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    You can just change everything to sin(x) and cos(x), then go from there.
    So:
    [tex]\int \frac{\sec ^ 4 x}{\tan ^ 2 x} dx = \int \frac{\cos ^ 2 x}{\sin ^ 2 x \cos ^ 4 x}dx = \int \frac{1}{\sin ^ 2 x \cos ^ 2 x}dx[/tex]
    There's sin(x) and cos(x) in the denominator, so you may want to change the numerator a bit so that it also has sin(x) and cos(x).
    [tex]... = \int \frac{\cos ^ 2 x + \sin ^ 2 x}{\sin ^ 2 x \cos ^ 2 x}dx = ...[/tex]
    Can you go from here?
    --------------------
    Or if you want to continue doing it your way, then you can change everything into tan(x).
    So:
    [tex]\int \frac{\sec ^ 4 x}{\tan ^ 2 x} dx = \int \frac{(1 + \tan ^ 2 x) ^ 2}{\tan ^ 2 x} dx = \int \frac{dx}{\tan ^ 2 x} + \int \tan ^ 2 x dx + \int 2 dx[/tex]
    Now you can change tan(x) to sin(x) and cos(x): tan(x) = sin(x) / cos(x). And use sin2x + cos2x = 1 to solve the problem.
    ---------------
    Anyway, it's always good that you start another thread for another problem, instead of using the old one...
    Viet Dao,
     
    Last edited: Oct 23, 2005
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