# How do you integrate 36/(2x+1)^3?

$- \frac{9}{2 x + 1} ^ 2 + C$
$u = 2 x + 1 R i g h t a r r o w \mathrm{du} = 2 \cdot \mathrm{dx}$
$\int \frac{36}{u} ^ 3 \cdot \frac{\mathrm{du}}{2} = 18 \int {u}^{- 3} \cdot \mathrm{du}$
$= 18 {u}^{- 2} / \left\{- 2\right\} + C = - \frac{9}{u} ^ 2 + C$