Evalute the integral:
((1-x)/x)^2 dx
I think I need to do integration by parts by I don't know the steps... please help
Thanks.
No need for integration by parts.
Recall that $\displaystyle \frac{1 - x}{x} = \frac{1}{x} - 1 = x^{-1} - 1$.
So $\displaystyle \left(\frac{1 - x}{x}\right)^2 = (x^{-1} - 1)^2$
$\displaystyle = x^{-2} - 2x^{-1} + 1$.
Therefore
$\displaystyle \int{\left(\frac{1 - x}{x}\right)^2\,dx} = \int{x^{-2} - 2x^{-1} + 1\,dx}$
$\displaystyle = -x^{-1} - 2\ln{|x|} + x + C$
$\displaystyle = x - \frac{1}{x} - 2\ln{x} + C$.