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  1. Default Integration by parts


    Integrate: 4p5ln(p)dp.
    I set u as 4p^5, and dv as ln(p). As I'm integrating, I'm noticing that I'll be using integration by parts on lnp over and over again. After noticing this, I'm unsure about if I'm doing this problem right, or this problem is really that horrible.

    Did I declare u and dv right, and is there no other way to get all the dv values other than integrating plnp 6 times?
    Last edited by Imagine; 2012-12-10 at 08:49 PM.

  2. Default Re: Integration by parts




    Should be right, I didn't really simplify anything though.

  3. Default Re: Integration by parts


    Wow, I can't believe that I forgot about using lnp as u.
    Thanks!

  4. Default Re: Integration by parts


    Bumping because I don't want to start a new thread.
    Integrate:
    3/((x + 1)(x^2 + 9))
    What I got was:
    (3/10)lnabs(x-1) - (3/20)(lnabs(x^2 + 9)) - (3/10)arctan(x/3) + C
    but it's wrong, and I can't seem to pinpoint was I messed up on.

  5. Default Re: Integration by parts


    Source




    Should be right.

  6. Default Re: Integration by parts


    stuck on another problem.
    Integrate:
    1/(x * sqrt(x + 1))

    Used u sub setting:
    u = sqrt(x +1)
    x = u^2 - 1
    dx = 2udu

    I end up with: -16/(u * (u^2 - 1))
    Move out -16, and split u^2 -1 into (u + 1)(u - 1)
    Used partial fractions; got A = -1, B = 0.5, C = 0.5.

    I get: -lnabs(u) + 1/2lnabs(u -1) + 1/2lnabs(u + 1) + C
    Substitute x back in, multiply by the -16 I moved out, and I get:
    16lnabs(sqrt(x + 1) - 8lnabs(sqrt(x + 1) - 1) - 8lnabs(sqrt(x+1) +1) + C
    And that seems to be wrong.

    Found my error, forgot to factor out u.
    Last edited by Imagine; 2012-12-10 at 08:49 PM.

  7. Default Re: Integration by parts


    Bumping because I have a math midterm and arc lengths suck.Find the length of lnx from 1 to sqrt(3).

    I'm getting u + 0.5ln(u + 1) - 0.5ln(u -1), where u = sqrt(1+ x^2), as the integral I'm plugging numbers into, but yahoo answer tells me it's sqrt(1 + x^2) - ln | [1 + sqrt(1 + x^2)] / x, so I dunno if I'm doing it right.

  8. Default Re: Integration by parts


    My first time working with arc lengths... which method are you using? I tried the standard method of Int{sqrt[1 + (dy/dx)^2]}, and it came out horribly. Had to use WolframAlpha to get that to work. Final output is: 2 - sqrt(2) + ln{[sqrt(2) + 1] / sqrt(3) }.

    Hadriel

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