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    how do i integrate this:

    2/(e^2x + 4) using u = denominator

    thanks
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    i think its this, but best wait for someone else to do it...

    Int 2/(e^2x + 4) dx
    u = e^2x + 4 => du/dx = 2e^2x

    Int 2/u du * dx/du = 2ln[u]/2e^2x + c
    2ln(e^2x + 4)/2e^2x + c
    = ln(e^2x + 4)/e^2x + c
    = [e^-2x](2x + ln4) + c
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    I = INT 2/(e^2x+4) dx.
    Using u=e^2x+4, du/dx=2e^2x, dx/du=0.5e^-2x
    I = INT (2/u).(0.5e^-2x) du
    I = INT (1/u).(e^-2x) du
    But e^-2x = 1/(u-4)
    I = INT (1/u).(1/u-4) du
    I = INT 1/u(u-4) du
    1/u(u-4) = A/u + B/u-4
    1=A(u-4) + B(u)
    u=4 gives B=1/4. u=0 gives A=-1/4.
    I = (1/4)INT 1/(u-4) - 1/u du
    I = (1/4)[ln(u-4) - lnu]+C
    I = (1/4)ln[(u-4)/u)]+C
    I = (1/4)ln(e^2x)/(e^2x+4) + C
    I = (1/4)ln(e^2x)-(1/4)ln(e^2x+4) + C
    I = (1/2)ln(e^x)-(1/4)ln(e^2x+4) + C
    I = (1/2)x - (1/4)ln(e^2x+4) +C
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    the book has answer 1/2x - 1/4 ln(u) + c

    i get close:

    u = e^2x + 4
    du/dx = 2e^2x
    dx = 1/(2e^2x) du

    so 2/u dx
    = (u - e^2x)/2/u dx
    = u - e^2x/2u dx
    = u - e^2x/ 2u .2e^2x du
    = u - e^2x/ 2u.2(u-4)
    = u - e^2x/ 4u^2 - 16u
    = u - (u - 4)/ 4u^2 - 16u
    = 4/4(u^2 - 4u)
    =1/u^2 - 4u
    = u^-2 - (u^-1)/4
    INT -u^-1 - 1/4 ln u
    = 1/e^2x +4 - 1/4 ln e^2x +4 + C
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    i think i see the problem - we need to spilt the fraction up from the beginning.
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    ok - i dont know - if any one can see how to get the solution to this let me know - thanks
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    (Original post by ryan750)
    ok - i dont know - if any one can see how to get the solution to this let me know - thanks
    I edited my post to show how it is equal to the answer in the book.
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    thanks gaz
 
 
 
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