`f(x)= -tanh^-1(sqrt(1-e^(2x)) + C` is not an incorrect solution.

The derivative of tanh^-1(x) is `1/(1-x^2)` .

The solution that you have proposed is just tanh^-1 expressed on the complex plane.

**Further Reading**

Step 1: integration by substitution

`u = 2x -gt du=2dx -gt dx=1/2du`

`= 1/2int(du)/(sqrt(1-e^u))`

Step 2: integration by substitution

`v = e^u -gt dv=e^udu -gt du=(dv)/v`

`= 1/2int(dv)/(vsqrt(1-v))`

Step 3: integration by substitution

`w = 1-v -gt v = 1-w`

`dw = -dv`

`=1/2int(dw)/((-1+w)sqrt(w))`

Step 4: integration by substitution

`z = sqrt(w) -gt w = z^2`

`dz = (dw)/(2sqrt(w)) -gt dw = 2zdz`

`= 1/2(2)int(zdz)/((-1+z^2)z)`

`= int(dz)/(-1+z^2)`

`= -int(dx)/(1-z^2)`

`= -tanh^-1(z) + C`

Final step: Resubstitute, w, v, and u:

`z=sqrt(w)=sqrt(1-v)=sqrt(1-e^u)=sqrt(1-e^2x)`

`int(dx)/(sqrt(1-e^(2x)))=-tanh^-1(sqrt(1-e^(2x)))+C`

**Further Reading**

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