What is the indefinite integral of f(x) = (e^x - 1)^1/2 ?

We'll write (e^x - 1)^1/2 = sqrt (e^x - 1). To evaluate
the indefinite integral, we'll substitute sqrt (e^x - 1) =
t.

We'll raise to square both
sides:

e^x - 1 = t^2

We'll add
1:

e^x = t^2 + 1

We'll take
logarithms both sides:

ln e^x = ln (t^2 +
1)

x*ln e = ln (t^2 + 1)

But
ln e = 1, so we'll get:

x = ln (t^2 +
1)

We'll differentiate both
sides:

dx = (t^2 + 1)'dt/(t^2 +
1)

dx = 2tdt/(t^2 +
1)

We'll evaluate the indefinite
integral:

Int sqrt(e^x-1)dx = Int t*2tdt/(t^2 +
1)

Int 2t^2dt/(t^2 + 1) = 2Int t^2dt/(t^2 +
1)

We'll add and subtract 1 to the
numerator:

 2Int (t^2 + 1 - 1)dt/(t^2 + 1) =  2Int
(t^2+1)dt/(t^2 + 1) - 2Int dt/(t^2 + 1)

We'll simplify and
we'll get:

 2Int (t^2 + 1 - 1)dt/(t^2 + 1) = 2Int dt -
2arctan t

We'll put t = sqrt (e^x - 1) and we'll
get:

Int sqrt(e^x-1)dx = 2sqrt(e^x-1) -
2arctan [sqrt(e^x-1)] + C