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Math Help - Tricky integral using substitution

  1. #1
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    Tricky integral using substitution

    The Problem
    Use the substitution u = 2x - 1 to find the integral of f(x) = x\sqrt{2x - 1}

    Attempt
    \frac{du}{dx} = 2x, so dx = \frac{1}{2} du.
    Therefore, this integral simplifies to \frac{1}{2}(Int\ x\sqrt{u}\ du).
    Rearranging, we get x = \frac{u + 1}{2}, so this becomes \frac{1}{2} x \frac{1}{2}\ Int (\sqrt{u}\ (u+1))\ du.
    Expanding, this becomes \frac{1}{4}\ Int\ (u^\frac{3}{2}\ + \sqrt{u})\ du.
    As everything is now a power of u, we can integrate as follows:
    \frac{1}{4}(\frac{u^\frac{5}{2}}{\frac{5}{2}}\ + \ \frac{u^\frac{3}{2}}{\frac{3}{2}}) .
    Multiplying out the fractions:
    \frac{1}{4}(\frac{2u^\frac{5}{2}}{5}\ +\ \frac{2u^\frac{3}{2}}{3}})
    Multiplying out the brackets:
    \frac{2u^\frac{5}{2}}{20}\ +\ \frac{2u^\frac{3}{2}}{12}
    Cancelling:
    \frac{u^\frac{5}{2}}{10}\ + \ \frac{u^\frac{3}{2}}{6}
    And as u = 2x - 1, we get:
    \frac{1}{10}(2x\ -\ 1)^\frac{5}{2} + \frac{1}{6}(2x\ -\ 1)^\frac{3}{2}+ C.

    I've checked this using 'Show Steps' after inputting the integral into WolframAlpha, and it appears to be correct.

    However, the answer given in both the textbook and WolframAlpha is:
    \frac{1}{15}(2x\ -\ 1)^\frac{3}{2}(3x\ +\ 1)+C.

    I would like a hint as to how I would progress from my answer to the answer given.
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  2. #2
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    The answers are equal.
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  3. #3
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    Try taking out \displaystyle (2x-1)^{\frac{3}{2}} as a common factor.
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  4. #4
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    Quote Originally Posted by TheCoffeeMachine View Post
    The answers are equal.
    Hey TCM, the OP has acknowledged this.
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  5. #5
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    Quote Originally Posted by pickslides View Post
    Hey TCM, the OP has acknowledged this.
    Oops, thanks! My apologies to the OP.
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  6. #6
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    OK, so what do I do next?

    OK, so I took out the common factor to get:
    (2x-1)^\frac{3}{2}(\frac{1}{10}(2x\ - 1)\ + \frac{1}{6}), then I could take out the factor of \frac{1}{2} to get:


    \frac{1}{2}(2x\ -1)^\frac{3}{2}(\frac{1}{5}(2x\ -1)\ + \frac{1}{3})
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  7. #7
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    Quote Originally Posted by TheCoffeeMachine View Post
    Oops, thanks! My apologies to the OP.
    That's fine.
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  8. #8
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    Quote Originally Posted by isx99 View Post
    OK, so I took out the common factor to get:
    (2x-1)^\frac{3}{2}(\frac{1}{10}(2x\ - 1)\ + \frac{1}{6}), then I could take out the factor of \frac{1}{2} to get:


    \frac{1}{2}(2x\ -1)^\frac{3}{2}(\frac{1}{5}(2x\ -1)\ + \frac{1}{3})
    Multiply the \frac{1}{10} into the brackets, group like terms, simplitfy and then take back out \frac{1}{15} you'll end up with the desired answer
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  9. #9
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    Thanks very much

    Quote Originally Posted by pickslides View Post
    Multiply the \frac{1}{10} into the brackets, group like terms, simplitfy and then take back out \frac{1}{15} you'll end up with the desired answer
    (That should be simplify, not simplitfy)

    Thanks very much. Just to confirm, would this be:

    (Expanding the brackets)
    (2x\ -1)^\frac{3}{2}(\frac{2x}{10}\ -\ \frac{1}{16}\ +\ \frac{1}{6}) \\ = \ (2x\ -1)^\frac{3}{2}(\frac{1}{5}x\ -\ \frac{1}{10}\ +\ \frac{1}{6}).

    (Adding the fractions)
    -\frac{1}{10}\ +\ \frac{1}{6} = \frac{4}{60}\ =\ \frac{1}{15}.

    (Simplifying)
    (2x\ -1)^\frac{3}{2}(\frac{1}{5}x\ -\ \frac{1}{15}).

    (Taking out the \frac{1}{15})
    \frac{1}{15}(2x\ -\ 1)^\frac{3}{2}(3x\ -\ 1), which is the answer given .
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