dx/3((2x-1)^2)
How do you integrate by substitution?
$\displaystyle \int \frac{dx}{3(2x-1)^2} $
$\displaystyle u = 2x-1 $
$\displaystyle \frac{du}{dx} = 2 $
$\displaystyle \frac{1}{2} du = dx $
$\displaystyle \int \frac{\frac{1}{2}du}{3u^2} $
$\displaystyle \frac{1}{2} . \frac{1}{3}\int \frac{du}{u^2} $
$\displaystyle \frac{1}{6} \int u^{-2}du $