Results 1 to 3 of 3

Thread: Integration by substitution, what am I doing wrong

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    631

    Integration by substitution, what am I doing wrong

    Using $\displaystyle u = \sqrt{2x+1} $, show that

    $\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c $

    I have done this so many times and keep getting the wrong answer, heres my working.

    $\displaystyle u = \sqrt{2x+1} $

    $\displaystyle x = \frac{1}{2}(u^{2} -1) $

    $\displaystyle \frac{dx}{du} = u $

    $\displaystyle dx = u du $

    $\displaystyle \int \frac{1}{2}(u^{2} -1) u u du $

    $\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du $

    $\displaystyle \frac{1}{2}\int (\frac{u^{5}}{5} - \frac{u^{3}}{3}) $

    $\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} ) $

    $\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3}) $

    From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by Tweety View Post
    Using $\displaystyle u = \sqrt{2x+1} $, show that

    $\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c $

    I have done this so many times and keep getting the wrong answer, heres my working.

    $\displaystyle u = \sqrt{2x+1} $

    $\displaystyle x = \frac{1}{2}(u^{2} -1) $

    $\displaystyle \frac{dx}{du} = u $

    $\displaystyle dx = u du $

    $\displaystyle \int \frac{1}{2}(u^{2} -1) u u du $

    $\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du $

    $\displaystyle \frac{u^{5}}{5} - \frac{u^{3}}{3} $

    $\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} ) $

    $\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3}) $

    From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

    thanks
    By factoring out $\displaystyle u^3$ in the last line we have

    $\displaystyle \frac{1}{30} (u^3)(3u^2-5) $

    Since $\displaystyle u=\sqrt{2x+1}$

    $\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 3 (2x+1) - 5) $

    $\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 6x+ 3-5)$

    $\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 6x-2) $

    $\displaystyle =\frac{1}{30} \sqrt{2x+1}^3 [2( 3x-1 ) ]$

    $\displaystyle = \frac{1}{15}\sqrt{2x+1}^3 ( 3x-1 ) $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by Tweety View Post
    Using $\displaystyle u = \sqrt{2x+1} $, show that

    $\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c $

    I have done this so many times and keep getting the wrong answer, heres my working.

    $\displaystyle u = \sqrt{2x+1} $

    $\displaystyle x = \frac{1}{2}(u^{2} -1) $

    $\displaystyle \frac{dx}{du} = u $

    $\displaystyle dx = u du $

    $\displaystyle \int \frac{1}{2}(u^{2} -1) u u du $

    $\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du $

    $\displaystyle \frac{1}{2}\int (\frac{u^{5}}{5} - \frac{u^{3}}{3}) $

    $\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} ) $

    $\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3}) $

    From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

    thanks
    $\displaystyle \int{x\sqrt{2x + 1}\,dx}$

    Let $\displaystyle u = 2x + 1$, then $\displaystyle x = \frac{u - 1}{2}$ and $\displaystyle \frac{du}{dx} = 2$.

    Go from here. Remember that $\displaystyle \sqrt{u} = u^{\frac{1}{2}}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Integration goes wrong? ( ln(x)/x )
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Mar 10th 2011, 10:34 AM
  2. integration what am i doing wrong?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 19th 2011, 03:16 AM
  3. What did I do wrong? (integration)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Aug 30th 2010, 05:07 AM
  4. Replies: 2
    Last Post: Jun 2nd 2010, 12:01 AM
  5. What is wrong with my Integration?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 14th 2010, 01:21 PM

Search Tags


/mathhelpforum @mathhelpforum