Results 1 to 8 of 8

Math Help - Integration by Parts!

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    40

    Integration by Parts!

    I have trouble with both problems:

    1. \int ln(x+3)dx

    2. \int (x-1)sinxdx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear littlesohi,

    1) \int{ln(x+3)dx}=(x+3)ln(x+3)-(x+3)

    2) \int{(x-1)Sinxdx}

    \int{xSinxdx}-\int{Sinxdx}

    -xCosx+\int{Cosxdx}+Cosx+C\mbox{ where C is an arbitary constant.}

    -xCosx+Sinx+Cosx+C

    Hope this helps.
    Last edited by Sudharaka; January 21st 2010 at 05:45 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    40
    for the one what is the u? dv?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear littlesohi,

    I didn't use integration by parts in the first problem. But you could use the integration by parts method as well.

    \int{ln(x+3)dx}

    \int{\frac{dx}{dx}\ln(x+3)dx}

    xln(x+3)-\int{\frac{x}{x+3}dx}

    I think you could do it from here.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    Quote Originally Posted by littlesohi View Post
    for the one what is the u? dv?
    u=\ln(x+3) ~\implies~ du=\frac{dx}{x+3}

    dv=dx ~\implies~ v=x


    Giving the result:

    uv - \int v \; du ~=~ x \ln(x+3) - \int \frac{x}{x+3} \; dx

    This is what Sudharaka got as well, but I think this is the way you are more used to seeing it done. You would need to use a substitution to finish the calculation.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Integration by parts is derived from the
    Product Rule.

    (not the Chain Rule as i originally wrote for unknown reasons!!)

    \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}


    v\frac{du}{dx}=\frac{d}{dx}uv-u\frac{dv}{dx}

    \int{v\frac{du}{dx}dx}=\int{\frac{d}{dx}(uv)dx}-\int{u\frac{dv}{dx}dx}

    \int{udv}=\int{d(uv)}-\int{vdu}

    \int{vdu}=uv-\int{udv}

    or

    \int{udv}=uv-\int{vdu}


    \int{ln(x+3)dx}=\int{udv}

    u=ln(x+3),\ dv=dx,\ v=x

    u=ln(x+3),\ \frac{du}{dx}=[\frac{1}{x+3}\ \frac{d}{dx}(x+3)]=\frac{1}{x+3}

    du=\frac{1}{x+3}dx

    uv-\int{vdu}=xln(x+3)-\int{\frac{x}{x+3}}dx
    Last edited by Archie Meade; January 22nd 2010 at 03:21 AM. Reason: massive typo!!!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Archie Meade View Post
    Integration by parts is derived from the
    Chain Rule.

    \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}
    It's a pity but that is the product rule.

    CB
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Yes! it certainly is, Captain!
    That's my greatest typo yet!!
    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: 1
    Last Post: January 24th 2010, 04:31 PM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum