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integrate

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integrate du/3u

and integrate

x/x^2 dx

it might be simple...i just need a head start...tnx

  • integrate -

    do i use the subsitution rule?

  • integrate -

    to be more specific...i know its best for me to attempt the problem

    so for the first problem i used substitution as
    v=3u
    dv=3du
    du=dv/3

    so answer i got was 1/3 ln3u

    is that correct?

    and on the second on i tried

    x*x^-2 as x^-3 dx so
    ans is -1/2x^-2

    is that correct? please let me know before i go ahead...i have a test tommorow and i am trying to do as many practice problem as i can...tnx

  • integrate -

    du/3u = (1/3) du/u
    which is (1/3)[ ln u + ln constant ]
    --> (1/3) ln ( c u)
    c = 3 works
    so
    (1/3) ln 3u + constant

    x/x^2 = 1/ x
    so
    dx/x --> ln x + constant

    I think x^1*x^-2 = x^-1 = 1/x

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