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# integration using substitution watch

1. how do i integrate this:

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

thanks
2. 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
3. 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
4. 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
5. i think i see the problem - we need to spilt the fraction up from the beginning.
6. ok - i dont know - if any one can see how to get the solution to this let me know - thanks
7. (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.
8. thanks gaz

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