How do I compute this improper integral?

Question

from 0 to 1 of (1/(square root{1-x}))dx

Thank you

Answer 1

u = 1-x
du = -dx
dx = -du
1/(√(1-x)) = 1/√u = u^(-½)

∫u^(-½)(-du) = -∫u^(-½)du = -[2u^½]

=-[2√(1-x)]

evaluated from 0 to 1:

=-[2√(1-1)] - -[2√(1-0)] = 2

Answer 2

This can be written as integral from 0 to 1 of (1+x)^ (-1/2) because the square root is 1/2 power and negative because it's in the denominator. Now add a power and divide by the new power to get (1+x) ^ (1/2) / (1/2) but dividing by 1/2 is the same as multiplying by 2. So you get 2 * (1+x) ^ (1/2) = 2 * sqrt(1+x). Plug in 1, subtract, plug in 0. 2 * sqrt(2) - 2 * sqrt(1)= 2* sqrt(2) -2