Calculus Question - Integration

I'd really like to know your thought process as to how to begin solving this problem, whether any substitution or any special technique, such as integration by parts, etc. were involved.

Thanks again!

Wait, I'm partially confused. :)

(1-x)/x is positive only when x is in (0, 1), so the sqrt is real only in the same range. Still, sqrt(x-1) is imaginary in this range.

sqrt((1-x)/x) (x-log(sqrt(x-1)+sqrt(x))sqrt(x/(x-1))

= x sqrt((1-x)/x) - i log(sqrt(x-1)+sqrt(x))
                          ^
                  hmm, indeed

log(ai+b) = log(sqrt(a^2+b^2)) + i*atan2(a,b)

Assuming 0 < x < 1
sqrt(x-1) = i*sqrt(1-x)

a=sqrt(1-x), b = sqrt(x)
log(sqrt(1-x)*i + sqrt(x)) = log(sqrt(1-x + x)) + i*theta
         = log(sqrt(1)) + i*theta
         = i*theta,

So it looks like the log term is purely imaginary when 0 < x < 1, so multiplying it by i makes it real.
given that 0 < x < 1,
theta = asin(sqrt(1-x) / sqrt(sqrt(1-x)^2 + sqrt(x)^2))
        = asin(sqrt(1-x) / sqrt(1 - x + x))
        = asin(sqrt(1-x))

So, it looks like we can replace the i*log with asin(sqrt(1-x))

integral = x*sqrt((1-x)/x) - asin(sqrt(1-x))
Let's take the derivitive and see what we get.

d{sqrt((1-x)/x)) = d{sqrt(1-x) / sqrt(x) } =
     (sqrt(x)*(-0.5/sqrt(1-x)) - sqrt(1-x)*(0.5/sqrt(x))) / x  =
multiply num and denom by sqrt(x)*sqrt(1-x)
     (x*(-0.5) - (1-x)*(0.5))/(x*sqrt(x)*sqrt(1-x)) =
     (-x/2 - 1/2 + x/2) =
     1/(2*x*sqrt(x)*sqrt(1-x))

d{x*sqrt((1-x)/x)} = 1/(2*sqrt(x)*sqrt(1-x)) + sqrt((1-x)/x)

d{asin(f(x))} = f'(x) * 1/sqrt(1 - f(x)^2)
d{asin(sqrt(1-x))} = (-0.5/sqrt(1-x)) * 1/sqrt(1 - (1-x))
                          = (-0.5/sqrt(1-x)) / sqrt(x)
                          = - 1/(2*sqrt(x)*sqrt(1-x))
adding it all together:
d{x*sqrt((1-x)/x) - asin(sqrt(1-x))}
     = 1/(2*sqrt(x)*sqrt(1-x)) + sqrt((1-x)/x) - 1/(2*sqrt(x)*sqrt(1-x))
     = sqrt(1-x)/x

So I believe that shows that the integral of sqrt(1-x)/x is x*sqrt((1-x)/x) - asin(sqrt(1-x)) + C

hello
it is very easy: look!
integral : ( x^2 - 2*x +1 ) / (x^2) dx
now divide ( x^2 - 2*x +1 ) to (x^2) . so the question will have another form:
integral : [ 1 - 2/x + 1/(x^2) ] dx
well, now we can easily solve it:
equals with : x - 2*Ln(x) - a/x + Constant
or
equals with : x - Ln(x^2) - a/x + Constant
the both are correct, no difference. Regards!

hello
it is very easy: look!
integral : ( x^2 - 2*x +1 ) / (x^2) dx
now divide ( x^2 - 2*x +1 ) to (x^2) . so the question will have another form:
integral : [ 1 - 2/x + 1/(x^2) ] dx
well, now we can easily solve it:
equals with : x - 2*Ln(x) - a/x + Constant
or
equals with : x - Ln(x^2) - a/x + Constant
the both are correct, no difference. Regards!

What are you on, CS?
>> integral : ( x^2 - 2*x +1 ) / (x^2) dx
This is not what we're trying to integrate;
this is:
>>Integrate sqrt((1-x)/x)dx

OOPS... ? should have been ² (square)....