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Math Help - Integrals

  1. #1
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    Integrals

    Can anyone help me out with these integrals step by step.

    The integral from 0 to 1 of (sqrt(t)) / (t + 1)

    Integral of (3x+2)/ x((x+2)^2)+16x

    any help would be appreciated thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by poopforbrains View Post
    The integral from 0 to 1 of (sqrt(t)) / (t + 1)
    \int_0^1 \frac{\sqrt{t}}{t + 1}~dt

    Let t = x^2 \implies dt = 2x~dx:
    \int_0^1 \frac{\sqrt{t}}{t + 1}~dt = \int_0^1 \frac{\sqrt{x^2}}{x^2 + 1} \cdot 2x~dx

    = 2 \int_0^1 \frac{x^2}{x^2 + 1}~dx

    Can you work with this one now?

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by poopforbrains View Post
    Integral of (3x+2)/ x((x+2)^2)+16x
    I would like you to verify. This integral is correct?
    \int \frac{3x + 2}{x(x + 2)^2 + 16x}~dx

    -Dan
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  4. #4
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    yup
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  5. #5
    Forum Admin topsquark's Avatar
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    \int \frac{3x + 2}{x(x + 2)^2 + 16x}~dx

    = \int \frac{3x + 2}{x^3 +4x^2 + 20x}~dx

    = \int \frac{3x + 2}{x(x^2 + 4x + 20)}~dx

    Now do partial fractions:
    \frac{3x + 2}{x(x^2 + 4x + 20)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4x + 20}

    (The kid's crying. Gotta go!)

    -Dan
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  6. #6
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    mmm I got the 2nd one but the first is still giving me trouble....I dont know if I should substitue again
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  7. #7
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    You don't need to set any substitution.

    Note that

    \frac{{x^2 }}{{x^2 + 1}} = \frac{{x^2 + 1 - 1}}{{x^2 + 1}} = 1 - \frac{1}{{x^2 + 1}}.

    When integrating, the first term it's just x, and the second one is the standard arctangent form.
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