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Thread: finding the integral

  1. #1
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    finding the integral

    Find the integral of:

    x * sqrt(5x^2 - 4) * dx
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  2. #2
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    Let $\displaystyle u = 5x^2 - 4$

    $\displaystyle du = 10x dx$

    OR

    $\displaystyle x dx = \frac{du}{10}$
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  3. #3
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    would it be:

    1/10 * (5x^2 - 4)^(3/2) + c
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    GAMMA Mathematics
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    Quote Originally Posted by DINOCALC09 View Post
    would it be:

    1/10 * (5x^2 - 4)^(3/2) + c
    Almost, you forgot to divide by the three halves.
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  5. #5
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    Quote Originally Posted by DINOCALC09 View Post
    Find the integral of:

    x * sqrt(5x^2 - 4) * dx
    Make-up

    $\displaystyle \int {x\sqrt {5x^2 - 4} \,dx} = \frac{1}
    {5}\int {5x\sqrt {5x^2 - 4} \,dx} .$

    Set $\displaystyle u^2=5x^2-4\implies u\,du=5x\,dx,$

    $\displaystyle \int {x\sqrt {5x^2 - 4} \,dx} = \frac{1}
    {5}\int {u^2 \,du} = \frac{u^3 }
    {{15}} + k = \frac{{\left( {5x^2 - 4} \right)^{3/2} }}
    {{15}} + k.$
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