What is the inverse of the function 2x-e^2x?

We'll put f(x) = 2x -
e^2x.

According to the rule, f'(x)*[f^-1(x)]' =
1

[f^-1(x)]'= 1/f'(x)

f^-1(x)
= Indefinite Integral of 1/f'(x)

We'll calculate f'(x) =
(2x-e^2x)'

f'(x) = 2 -
2e^2x

[f^-1(x)]'= 1/(2 -
2e^2x)

We'll calculate the indefinite
integral:

Int dx/(2 - 2e^2x) = (1/2)*Int dx/(1 -
e^2x)

We'll put 1 - e^2x = t => e^2x = 1 -
t

We'll
differentiate:

2e^2x*dx =
-dt

dx = -dt/2e^2x

dx =
-dt/2(1 - t)

(1/2)*Int dx/(1 - e^2x) = (1/2)*Int -dt/2t*(1
- t)

(1/2)*Int -dt/2t*(1 - t) = (-1/4)*Int dt/t*(1 -
t)

We'll decompose the fraction 1/t*(1-t) in a sum or
difference of elementary fractions:

1/t*(1 - t) = A/t +
B/(1-t)

1 = A - At + Bt

1 =
t(B-A) + A

B-A = 0

A =
B

A = 1 => B = 1

1/t*(1
- t) = 1/t + 1/(1-t)

(-1/4)*Int dt/t*(1 - t) = (-1/4)*[Int
dt/t + Int dt/(1 - t)]

(-1/4)*Int dt/t*(1 - t) = (-1/4) [ln
|t| + ln |1 - t|]  +C

(-1/4)*Int dt/t*(1 - t) = (-1/4) ln
|t*(1-t)| + C

Int dx/(2 - 2e^2x)  = (-1/4) ln |(1 -
e^2x)*(e^2x)| + C

The inverse function is
f^-1(x) = -[ln |(1 - e^2x)*(e^2x)|]/4.