I bailed out after whole page of messy attempts using integration by parts.
http://integrals.wolfram.com/index.jsp?expr=1%2F%28x%281-x%5E%281%2F2%29%29%5E2%29&random=false
Wolfram's answer assumes that log x is ln(x)
Calculate the following integral:
∫ dx / x(1 - √x)^2
3 answers
hmmm. Doesn't look too bad.
u = 1-√x
x = (1-u)^2
dx = -2(1-u) du
dx/(x(1-√x)^2)
= -2(1-u) du / ((1-u)^2 * u^2)
= -2 /((1-u)*u^2) du
= -2/u^2 -2/u + 2/(u-1) du
= 2(1/u + ln(1-u) - ln(u))
Now substitute u = 1-√x to get back to x
u = 1-√x
x = (1-u)^2
dx = -2(1-u) du
dx/(x(1-√x)^2)
= -2(1-u) du / ((1-u)^2 * u^2)
= -2 /((1-u)*u^2) du
= -2/u^2 -2/u + 2/(u-1) du
= 2(1/u + ln(1-u) - ln(u))
Now substitute u = 1-√x to get back to x
Thanks Reiny and Steve. Now to pass the exam!