Results 1 to 3 of 3

Math Help - Integration of Rational Expression - Alternative Method

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    74

    Integration of Rational Expression - Alternative Method

    Hello everyone,

    I integrated the following rational expression correctly, but one of my steps differ from that one of the book. Could anyone tell me if what I did is still correct?

    Thank you!

    ---

    1. Integrate:  \int \frac{\sqrt{x - 4} + x}{x - 4} dx

    ---
     \int \frac{\sqrt{x - 4} + x}{x - 4} dx

    =  \int (x - 4)^{-1/2} dx + \int \frac{x}{x - 4}dx

    Let  u = x - 4 and  du = dx . Therefore,  x = u + 4.

    =  \int u^{-1/2} du + \int \frac{u + 4}{u}du

    =  2\sqrt{x - 4} + \int \frac{u}{u} du + \int \frac{4}{u} du

    =  2\sqrt{x - 4} + x + \ln(x - 4)^4 + C

    ---

    However, my practice workbook integrated  \int \frac{u}{u} du as  u , and then substituted  x - 4 back into the equation.

    So according to the workbook, its last step would be:

    =  2\sqrt{x - 4} + u + \ln(x - 4)^4 + C

    =  2\sqrt{x - 4} + (x - 4) + \ln(x - 4)^4 + C

    =  2\sqrt{x - 4} + x + \ln(x - 4)^4 + C
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    I think the way you did it kind of implies that \int \frac{u}{u}du=x because of the way you kept your terms in order. The book's method is more correct in my opinion as it notes that the above integral doesn't equal x, but u + C, then back-substitutes x-4 for u and combines all the constants into one C.
    Last edited by Jameson; August 5th 2008 at 06:27 AM. Reason: small mistake
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608
    Hello, scherz0!

    What you did is correct.
    . . I would not have simplifed first as you did . . .


    We are given: . \int \frac{\sqrt{x - 4} + x}{x - 4}\,dx

    Let: \sqrt{x-4} \:=\: u \quad\Rightarrow\quad x-4\:=\:u^2 \quad\Rightarrow\quad x \:=\: u^2+4 \quad\Rightarrow\quad dx \:=\:2u\,du


    Substitute: . \int \frac{u + (u^2+4)}{u^2}\,(2u\,du) \;\;=\;\;2\int\frac{u^2+u-4}{u}\,du \;\;=\;\; 2\int\left(u + 1 - \frac{4}{u}\right)\,du

    . . . . . . = \;\;2\left(\frac{u^2}{2} + u - 4\ln|u|\right) + C \;\;=\;\;u^2 + 2u - 8 \ln|u| + C


    Back-substitute: . (x - 4) + 2\sqrt{x-4} - 8\ln\left(\sqrt{x-4}\right) + C

    . . . . . . . . . . . =\;\;x - 4 + 2\sqrt{x-4} - 8\ln(x-4)^{\frac{1}{2}} + C

    . . . . . . . . . . . = \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + \underbrace{C - 4}_{\text{a constant}}

    . . . . . . . . . . . = \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. alternative expression of L^2
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 8th 2011, 06:14 PM
  2. Alternative method for binomial denominators?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 13th 2010, 03:20 PM
  3. Replies: 2
    Last Post: December 8th 2009, 06:48 AM
  4. any alternative method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 11th 2009, 03:35 AM
  5. Rational Expression
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 9th 2008, 03:57 PM

Search Tags


/mathhelpforum @mathhelpforum