Results 1 to 3 of 3

Math Help - integration by parts help

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    integration by parts help

    \int x\tan^{-1}(x) dx

    let u = tan^-1(x)
    du = \frac{1}{1+x^2}dx

    let v = \frac{x^2}{2}
    dv = x

    then got stuck after this
    \frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \int x\tan^{-1}(x) dx

    let u = tan^-1(x)
    du = \frac{1}{1+x^2}dx

    let v = \frac{x^2}{2}
    dv = x

    then got stuck after this
    \frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx
    Doing great so far. Here's how to deal with the new integral.

    forget the constant 1/2 you can pull out.

    \int \frac {x^2 }{1 + x^2}~dx = \int \frac {x^2 + 1 - 1}{1 + x^2}~dx

    = \int \left( 1 - \frac {1}{1 + x^2} \right)~dx

    Now continue
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \int x\tan^{-1}(x) dx

    let u = tan^-1(x)
    du = \frac{1}{1+x^2}dx

    let v = \frac{x^2}{2}
    dv = x

    then got stuck after this
    \frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx
    Try this:
    \frac{x^2}{x^2 + 1} = \frac{x^2 + 1 - 1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum