Results 1 to 5 of 5

Math Help - Intrgration by Parts

  1. #1
    Junior Member
    Joined
    Sep 2011
    Posts
    27

    Intrgration by Parts

    Hi,

    I'm not sure where to begin with find the anti derivative of this function... I know it's IBP, but I'm really confused. Any help is appreciated.

    = \int_ (x^3\sqrt{2+x^2}dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Intrgration by Parts

    Do the substitution, let 2+x^2=t.
    Can you go further?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Intrgration by Parts

    If you are using integration by parts then...

    \int x^3\sqrt{2+x^2}dx

    =\frac{x^4\sqrt{2+x^2}}{4}-\frac{1}{4}\int \frac{x^5}{\sqrt{2+x^2}}dx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49

    Re: Intrgration by Parts

    Quote Originally Posted by sbhatnagar View Post
    If you are using integration by parts then...

    \int x^3\sqrt{2+x^2}dx

    =\frac{x^4\sqrt{2+x^2}}{4}-\frac{1}{4}\int \frac{x^5}{\sqrt{2+x^2}}dx
    That would be a useful learning experience... but afterwards, integrate the square root instead. Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    But this is wrapped inside the legs-uncrossed version of...



    ... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.



    ... is lazy integration by parts, doing without u and v.


    Spoiler:


    The sub instead...

    Spoiler:




    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; September 26th 2011 at 02:06 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644

    Re: Intrgration by Parts

    Hello, sjmiller!

    I \;=\; \int  x^3\sqrt{x^2+2}\,dx

    If we must integrate by parts . . .

    . . \begin{Bmatrix} u &=& x^2 && dv &=& x(x^2+2)^{\frac{1}{2}}dx \\ du &=& 2x\,dx ^&& v &=& \frac{1}{3}(x^2+2)^{\frac{3}{2}} \end{Bmatrix}


    \text{Then: }\:I \;=\;\tfrac{1}{3}x^2(x^2+2)^{\frac{3}{2}} - \tfrac{2}{3}\int x(x^2+2)^{\frac{3}{2}}dx

    . . . . . . . =\;\tfrac{1}{3}x^2(x^2+2)^{\frac{3}{2}} - \tfrac{2}{15}(x^2+2)^{\frac{5}{2}} + C

    . . . . . . . =\;\tfrac{1}{15}(x^2+2)^{\frac{3}{2}}\bigg[5x^2 - 2(x^2+2)\bigg] + C

    . . . . . . . =\;\tfrac{1}{15}(x^2+2)^{\frac{3}{2}}\left(3x^2-4\right) + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. int by parts
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 19th 2010, 05:21 PM
  2. Do I have to use By Parts?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 4th 2010, 05:59 AM
  3. Int. by parts (x-1)e^2x-1
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 15th 2009, 02:57 AM
  4. different int. by parts
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 22nd 2009, 02:08 PM
  5. PARTS HELP IM NEW HERE
    Posted in the Calculus Forum
    Replies: 13
    Last Post: September 26th 2008, 08:39 AM

Search Tags


/mathhelpforum @mathhelpforum