# Indefinite integral Determine the indefinite integral of f(x) = (e^x - 1)^1/2 ?

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You should come up with the following substitution, such that:

`e^x - 1 = u^2 => e^x dx = 2udu => dx = (2udu)/(1 + u^2) `

Changing the variable yields:

`int (e^x - 1)^(1/2)dx = int (u^2)^(1/2)(2udu)/(1 + u^2) `

`int 2u^2/(1 + u^2) du = 2 int u^2/(1 + u^2)du`

You need to use reminder theorem such that:

`u^2 = (u^2+1) - 1 => u^2/(1 + u^2) = 1 - 1/(1 + u^2)`

Integrating both sides the relation `u^2/(1 + u^2) = 1 - 1/(1 + u^2)` yields:

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

Using the property of linearity of integral yields:

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

`2int u^2/(1 + u^2) du = 2(u - tan^(-1) u) + c`

Substituting back `sqrt(e^x - 1)` for u yields:

`int (e^x - 1)^(1/2)dx = 2(sqrt(e^x - 1) - tan^(-1) sqrt(e^x - 1)) + c`

**Hence, evaluating the indefinite integral**`int (e^x - 1)^(1/2)dx yields int (e^x - 1)^(1/2)dx = 2(sqrt(e^x - 1) - tan^(-1) sqrt(e^x - 1)) + c.`

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**