Integrate e^2x/e^x 1....plzz note e^x 1 as a whole is the denominator
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?
To integrate the given function, e^(2x) / (e^x + 1), we can use a substitution method. Let's set u = e^x + 1, and then find the expression for du in terms of dx.
Differentiating both sides of the equation u = e^x + 1 with respect to x, we have:
du/dx = d/dx(e^x + 1)
Since the derivative of e^x with respect to x is simply e^x, the above equation simplifies to:
du/dx = e^x
Solving for dx, we have:
dx = du / e^x
Now let's substitute these values into the integral:
∫ (e^(2x) / (e^x + 1)) dx
= ∫ (e^(2x) / u) * (du / e^x)
= ∫ e^x * (e^x / u) du
Simplifying further, we get:
∫ e^x * (e^x / (e^x + 1)) du
= ∫ (e^x / (e^x + 1)) du
Next, we can let v = e^x + 1, where dv = e^x dx. We can rewrite the integral using v:
∫ (1 / v) dv
Now we can integrate this simplified integral:
∫ (1 / v) dv = ln|v| + C
Finally, substituting back the value of v:
ln|e^x + 1| + C
Therefore, the integral of e^(2x) / (e^x + 1) is ln|e^x + 1| + C, where C is the constant of integration.