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  1. #1
    sss
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    integration

    1.A particle that moves along a straight line has velocity
    meters per second after t seconds. How many meters will it travel during the first t seconds?

    I used integration by parts but i didnt get the write answer.

    2.






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  2. #2
    o_O
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    (1) You're going to need to use integration by parts twice.

    First round requires: u = t^2 and dv = e^{-2t} dt

    You will get: \int t^2 e^{-2t} = -\frac{1}{2}t^2e^{-2t} + 2 {\color{red}\int te^{-2t}dt}

    Use integration by parts again on the red: u = t and dv = e^{-2t}dt

    Can you carry on from here?

    ________________________

    (2) Again, you'll need integration by parts: u = \arctan (3x) and dv = xdx

    You'll get: \int x\arctan (3x) dx = \frac{x^2}{2}\arctan (3x) - \frac{3}{2} {\color{blue}\int \frac{x^2}{1+9x^2} dx}

    The blue integral should be easier to deal with now. You'll need long division or you can notice that: \frac{x^2}{1+9x^2}  \cdot  \frac{9}{9} = \frac{9x^2 \ {\color{red}+ 1 - 1}}{9 (1+9x^2)} = \frac{1}{9} - \frac{1}{9} \left( \frac{1}{1+9x^2}\right)
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by sss View Post
    I used integration by parts but i didnt get the write answer.
    2.

    \int {x\arctan \left( {3x} \right)} dx = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{1}{2}\int {\frac{{3{x^2}}}{{{{\left( {3x} \right)}^2} + 1}}dx}  =

    = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{1}{6}\int {\frac{{{{\left( {3x} \right)}^2} + 1 - 1}}{{{{\left( {3x} \right)}^2} + 1}}dx}  =

    = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{6}\int {\frac{{dx}}{{{{\left( {3x} \right)}^2} + 1}}}  =

    = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{{18}}\int {\frac{{d\left( {3x} \right)}}{{{{\left( {3x} \right)}^2} + 1}}}  =

    = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{{18}}\arctan \left( {3x} \right) + C =

    = \frac{1}{{18}}\left( {9{x^2} + 1} \right)\arctan \left( {3x} \right) - \frac{x}{6} + C.
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