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...

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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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