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    x\sqrt{2x+1} using  u=\sqrt{2x+1}

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

     \frac{u^{2}-1}{2}\times \frac{u^2}{1}

     \frac{1}{2}\int [ u^4-u^2 \]

    \frac{1}{2}\left [ \frac{3u^5-5u^3}{15} \right ]


    please help me?
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    EDIT: Your working seems correct.

    Just substitute back at the end.
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    (Original post by notnek)
    Where did \frac{u^2}{1} come from?
    It's not clear, but \frac{dx}{du}

    gives dx = u du

    so we get u squared in the transformed integral.
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    to start, dx=udu

    integrand becomes:

    \frac{u^{2}-1}{2} \times u
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    (Original post by Indeterminate)
    It's not clear, but \frac{dx}{du}

    gives dx = u du

    so we get u squared in the transformed integral.
    You're right - I noticed my mistake as soon as I posted.
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    (Original post by Hasufel)
    to start, dx=udu

    integrand becomes:

    \frac{u^{2}-1}{2} \times u

    rearrange the x and sub it into the question

    u = squareroot(2x+1)

    then sub the dx in, then times it all together.

    Sorry i didnt make it clear but typing all this in latex is frustrating since I had so many steps
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    (Original post by franzk)
    x\sqrt{2x+1} using  u=\sqrt{2x+1}

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

     \frac{u^{2}-1}{2}\times \frac{u^2}{1}

     \frac{1}{2}\int [ u^4-u^2 \]

    \frac{1}{2}\left [ \frac{3u^5-5u^3}{15} \right ]


    please help me?
    It's fine so far. Factorise:

    \displaystyle \frac{u^3}{30}\left( 3u^2-5 \right)

    Now sub u back in and remember the constant of integration.
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    Would it not be easier to use integration by parts?
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    (Original post by notnek)
    It's fine so far. Factorise:

    \displaystyle \frac{u^3}{30}\left( 3u^2-5 \right)

    Now sub u back in and remember the constant of integration.

    Thank you I managed to work it out, I just didnt expect so much factorising.. i even had to factorise further to get it into the form stated in the question.

    (Original post by thelion0)
    Would it not be easier to use integration by parts?

    question says do it by substitution.
 
 
 
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