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Math Help - Integration by substitution, what am I doing wrong

  1. #1
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    Integration by substitution, what am I doing wrong

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

     \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.

     u = \sqrt{2x+1}

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

     \frac{dx}{du} = u

     dx  = u du

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

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

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

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

     \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
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Using  u = \sqrt{2x+1} , show that

     \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.

     u = \sqrt{2x+1}

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

     \frac{dx}{du} = u

     dx = u du

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

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

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

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

     \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  u^3 in the last line we have

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

    Since  u=\sqrt{2x+1}

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

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

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

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

     =  \frac{1}{15}\sqrt{2x+1}^3 ( 3x-1 )
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  3. #3
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    Quote Originally Posted by Tweety View Post
    Using   u = \sqrt{2x+1} , show that

     \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.

     u = \sqrt{2x+1}

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

     \frac{dx}{du} = u

     dx  = u du

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

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

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

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

     \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
    \int{x\sqrt{2x + 1}\,dx}

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

    Go from here. Remember that \sqrt{u} = u^{\frac{1}{2}}.
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