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

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.