We'll put f(x) = e^x+6x.
By
definition, f'(x)*[f^-1(x)]' = 1
[f^-1(x)]'=
1/f'(x)
f^-1(x) = Integral of
1/f'(x)
We'll calculate f'(x) = e^x +
6
[f^-1(x)]'= 1/(e^x +
6)
We'll calculate the indefinite
integral:
Int dx/(e^x +
6)
We'll put e^x + 6 = t => e^x = t -
6
We'll differentiate:
e^x*dx
= dt
dx = dt/e^x
dx =
dt/(t-6)
Int dx/(e^x + 6) = Int
dt/t*(t-6)
We'll decompose the fraction 1/t*(t-6) in a sum
or differenceof elementary fractions:
1/t*(t-6) = A/t +
B/(t-6)
1 = t(A+B) - 6A
A+B =
0
A = -B
A = -1/6 => B
= 1/6
1/t*(t-6) = -1/6t +
1/6(t-6)
Int dt/t*(t-6) = -Int dt/6t + Int
dt/6(t-6)
Int dt/t*(t-6) = (1/6)(Int dt/(t-6) - Int
dt/t)
Int dt/t*(t-6) = (1/6)(ln(t-6) - ln
t)
Int dt/t*(t-6) =
(1/6){ln[(t-6)/t]}
Int dx/(e^x + 6) = (1/6){ln[(e^x)/(e^x +
6)]} +C
The inverse function is f^-1(x) =
{ln[(e^x)/(e^x + 6)]}/6
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