ok, what is "e^x 1" ?
.Sorry its e^x+1.....:P
If it is e^(2x)/(e^x + 1) the way you typed it, then the integral is
e^c - ln(e^x + 1) + C
if you meant:
e^(2x)/e^(x+1) then it would simplify to
e^(x-1) and the integral of that is simply
e^(x-1) + c
Could u plzz tell it in detail...the 1st 1
let u = e^x+1.
du = e^x dx
and you have
∫(u-1)/u du = ∫ 1 - 1/u du = u - lnu = (e^x+1) - ln(e^x+1) + C
by absorbing the +1 into the C, you end up with
= e^x - ln(e^x+1) + C
...u see the numrtr is e^2x and how can u write it as u-1?
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