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Math Help - integration by parts

  1. #1
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    integration by parts

    use integration by parts to find each of the following integrals:

    \int tan^{-1} \left(\frac{x}{2}\right) dx

    \int x^2 e^{2x} dx

    \int log_3(2x+1) dx

    thanks : )
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  2. #2
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    Quote Originally Posted by qzno View Post
    use integration by parts to find each of the following integrals:

    \int tan^{-1} \left(\frac{x}{2}\right) dx


    u = \tan^{-1}\left(\frac{x}{2}\right) <br />

    du = \frac{2}{4+x^2} \, dx

    dv = dx

    v = x

    ---------------------------------------------------

    \int x^2 e^{2x} dx


    u = x^2

    du = 2x \, dx

    dv = e^{2x} \, dx

    v = \frac{1}{2}e^{2x}

    btw ... this one is easier if you use tabular integration.

    ----------------------------------------------------------

    \int log_3(2x+1) dx

    change base ...

    \frac{1}{\ln{3}} \int \ln(2x+1) dx


    u = \ln(2x+1)

    du = \frac{2}{2x+1} \, dx

    dv = dx

    v = x
    go for it.
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  3. #3
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    thank you : )
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  4. #4
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     xtan^{-1}\left(\frac{x}{2}\right) - \frac{x^2}{4} + ln x^2 + C
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  5. #5
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    i got something different:

    xtan^{-1}\left(\frac{x}{2}\right) - \frac{1}{4} + ln x^2 + C
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  6. #6
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    \int \tan^{-1}\left(\frac{x}{2}\right) \, dx<br />

    u = \tan^{-1}\left(\frac{x}{2}\right)

    du = \frac{2}{4 + x^2} \, dx<br />

    dv = dx

    v = x


    \int \tan^{-1}\left(\frac{x}{2}\right) \, dx = x \cdot \tan^{-1}\left(\frac{x}{2}\right) - \int \frac{2x}{4 + x^2} \, dx

    \int \tan^{-1}\left(\frac{x}{2}\right) \, dx = x \cdot \tan^{-1}\left(\frac{x}{2}\right) - \ln(4 + x^2) + C
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  7. #7
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    are the other two answers:

    \frac{1}{2}e^{2x} \left(x^2 - x + \frac{1}{2}\right) + C

    and

    \frac{1}{ln3}\left(xln(2x+1) - x - x^2\right) + C

    ???
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  8. #8
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    how does one check an antiderivative?
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  9. #9
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    take the derivative of the answer.
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