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Calculus:

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    Calculus:


    Integration by Substitution(reverse derivatives)
    1:

    x / (3x+2)^(1/2)

    2:

    1 / (5x-3)^(1/2)

    3:

    Sin4x

    4:

    (1+x^3)^1/2 * (x^2)

    5:

    (x^2+5)^3

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    you'll never use that in the real world....unless you grow up to be a calc teacher.
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    If I do these problems for you, how does that help you?

    Besides, "learning" calculus on a forum is not the best way to go about it.
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    By God, I sucked at calculus, but damnit I can figure your BMR in seconds.
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    Originally posted by chosen5
    Integration by Substitution(reverse derivatives)
    1:

    x / (3x+2)^(1/2)

    2:

    1 / (5x-3)^(1/2)

    3:

    Sin4x

    4:

    (1+x^3)^1/2 * (x^2)

    5:

    (x^2+5)^3
    1. [2*(3x-4) * (3x+2)^.5] / 27
    2. [2*(5x-3)^.5] / 5
    3. -45*cos(4x) / pi
    4. [2*(x^3+1)^3/2] / 9
    5. (x^7) / 7 + 3x^5 + 25x^3 + 125x


    There's your answers (quick and dirty) but the actual work is up to you.

    Chemo
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    hmm...chemo, can you do my taxes on here?
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