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

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Given the function:

f(x) = (e^x  -1)^1/2

==> intg f(x) = intg (e^x-1)^1/2 dx

Let us assume that u= (e^x -1)^/2

==> u^2 = e^x -1

==> e^x = u^2 +1

==> x = ln(u^2 +1)   ==> dx = 2u/(u^2+1) du

Now we will substitute:

==> intg f(x) = intg u  *2u du// (u^2+1)

==> intg f(x) = intg (2u^2 /(u^2+1)  du

= 2*intg (u^2/(u^2 +1)  du

= 2*intg (u^2+1-1) / (u^2+1)  du

= 2[intg (u^2+1)/(u^2+1) du - intg 1/(u^2+1)  du]

= 2[  intg du  - intg 1/(u^2+1) du

= 2( u - arctanu)  + C

= 2( e^x-1) - arctan(e^x -1)^1/2 + C

= 2e^(x-1) - 2arctan(sqrt(e^x-1) + C

==>intg f(x) = 2e^(x-1) - 2arctan(sqrt(e^x-1))+ C

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To find the  indefinite integral of f(x) = (e^x - 1)^1/2.

f(x) = (e^x-1)^1/2.

Put (e^x-1)^(1/2) = t.

Differentiating, we get:  e^x dx  /2(e^x-1)^(1/2) = t.

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

Also (e^x-1)^(1/2) = t. So e^x = t^2+1.

Therefore Int f(x) dx =  Int t *dx = Int 2t^2/(t^2+1)

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

= 2t - 2arc tan t.

= 2(e^x-1)^(1/2 - 2arc tan {(e^x-1)^(1/2)} + C.

Therefore Int (e^x - 1)^1/2 dx = 2(e^x-1)^(1/2 - 2arc tan {(e^x-1)^(1/2)} + C.

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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

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