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Math Help - antidifferentiation problem. is this correct?

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    antidifferentiation problem. is this correct?

    For the rule integral sign 1/x dx = ln lxl . does it only apply for 1/x? I mean does integral sign 1/2x dx = ln l2xl ?

    Also
    The problem is:

    integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

    Is this correct:

    integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

    = -3 x -e^(-x) + 2 x (1/2) x^2 - (1/2)(1/0.5)(e^(.5x)) + C
    =3(e^(-x))+x^2-(e^(.5x)) + C
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  2. #2
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    Quote Originally Posted by fabxx View Post
    For the rule integral sign 1/x dx = ln lxl . does it only apply for 1/x? I mean does integral sign 1/2x dx = ln l2xl ?

    Also
    The problem is:

    integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

    Is this correct:

    integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

    = -3 x -e^(-x) + 2 x (1/2) x^2 - (1/2)(1/0.5)(e^(.5x)) + C
    =3(e^(-x))+x^2-(e^(.5x)) + C

    \int \frac{1}{2x} \, dx = \frac{1}{2}\int \frac{1}{x} \, dx = \frac{1}{2} \ln|x| + C = \ln{\sqrt{x}} + C <br />


    please do not use "x" as the symbol for multiplication, especially when using x as a variable. common form is to use the asterisk (*).


    \int -3e^{-x} + 2x - \frac{1}{2}e^{\frac{x}{2}} \, dx = <br />

    3e^{-x} + x^2 - e^{\frac{x}{2}} + C

    remember you can check integrals by taking the derivative of your solution.
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