# Find the antiderivative of y= square root(e^x-1)?Find the antiderivative of y= square root(e^x-1)?

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### 1 Answer

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 calculate the antiderivative:

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

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