evaluate the integral:

integral of ((1-x)/x)^2 dx

Expand the integrand to:

1/x^2 -2/x + 1

The integral ia the sum of the integrals of each term

1/x - 2 lnx + x (+ any constant)

integral of ((1-x)/x)^2 dx

= integral of (1-2x+x^2)/(x^2) dx
= integral of (1/x^2 - 2/x + 1) dx
= -1/x - 2(lnx) + x + C

Reiny is correct. I missed a sign on one term. wls

To evaluate the integral of ((1-x)/x)^2 dx, we can expand the expression and then integrate each term separately. Let's walk through the steps:

Step 1: Expand the expression ((1-x)/x)^2:
((1-x)/x)^2 = (1-2x+x^2)/x^2 = 1/x^2 - 2/x + 1

Step 2: Integrate each term separately:
∫(1/x^2 - 2/x + 1) dx = ∫(1/x^2) dx - ∫(2/x) dx + ∫(1) dx

Step 3: Evaluate each integral:
∫(1/x^2) dx = -1/x
∫(2/x) dx = 2ln|x|
∫(1) dx = x

Step 4: Combine the results:
∫((1-x)/x)^2 dx = -1/x + 2ln|x| + x + C

Therefore, the integral of ((1-x)/x)^2 dx is equal to -1/x + 2ln|x| + x + C, where C is the constant of integration.