# Evaluate the integral of f(x)= square root x/(x-1)

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The integral of the function returns the primitive function F(x), such as dF/dx = f(x).

We'll evaluate the primitive function F(x), integrating the given function f(x):

Int f(x)dx = F(x) + C

Int sqrtx dx/(x-1)

We'll consider the denominator of the function as a difference of two squares that returns the product:

x - 1 = (sqrtx - 1)(sqrtx + 1)

We'll re-write the integral:

Int sqrtx dx/(sqrtx - 1)(sqrtx + 1)

We'll add and subtract 1 to the numerator:

Int (sqrtx + 1 - 1 )dx/(sqrtx - 1)(sqrtx + 1)

We'll re-group the terms of integrand using the property of integral of being additive:

Int (sqrtx + 1 - 1 )dx/(sqrtx - 1)(sqrtx + 1) = Int(sqrtx + 1)dx/(sqrtx - 1)(sqrtx + 1) - Int dx/(sqrtx - 1)(sqrtx + 1)

We'll simplify and we'll get:

Int (sqrtx + 1 - 1 )dx/(sqrtx - 1)(sqrtx + 1) = Int dx - Int dx/(x-1)

Int f(x)dx = x - ln|x-1| + C

**The indefinite integral of the function is: F(x) = x - ln|x-1| + C.**