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Math Help - Help with indefinite integrals?

  1. #1
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    Help with indefinite integrals?

    antiderivative [(3x)/((2x^2+1)^1/2)]
    what i did:
    let u=2x^2+1
    du/dx=4x
    (3/4)du=(3/4)(4x)(dx)
    3/4 du=(3x)(dx)
    x=[(u-1)/2)]^1/2
    antiderivative [(u^1/2)(3[(u-1)/2]^1/2)(du)]
    = antiderivative [(3/u^1/2)[(u-1)/2]^1/2
    from here i get stuck..i need to multiply du in as well but im not sure how to do that since i only found 3/4du=3xdx
    how would i do this? or am i doing something wrong?
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  2. #2
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    This looks OK to me

    \displaystyle \int \frac{3x}{\sqrt{2x^2+1}}~dx = \frac{3}{4}\int \frac{4x}{\sqrt{2x^2+1}}~dx =\frac{3}{4} \int \frac{1}{\sqrt{u}} = \frac{3}{4}\frac{\sqrt{u}}{\frac{1}{2}}+C=\dots
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  3. #3
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    Quote Originally Posted by pickslides View Post
    This looks OK to me

    \displaystyle \int \frac{3x}{\sqrt{2x^2+1}}~dx = \frac{3}{4}\int \frac{4x}{\sqrt{2x^2+1}}~dx =\frac{3}{4} \int \frac{1}{\sqrt{u}} = \frac{3}{4}\frac{\sqrt{u}}{\frac{1}{2}}+C=\dots
    Thank youu =]
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