$\int{\frac{8y}{4-y^2}dy}$
The answer isn't in the back of my book, so I have no way to see if I'm right! (I'm about 99% sure I'm wrong though)
$\int{\frac{8y}{4-y^2}dy}$
The answer isn't in the back of my book, so I have no way to see if I'm right! (I'm about 99% sure I'm wrong though)
$\int\frac{8y}{4-y^2}\,dy=-4\int\frac{d(4-y^2)}{4-y^2}=-4\log|4-y^2|+K\,\,(constant)$
Letting $u = 4- y^2$, we have
$\int \frac{8y}{4-y^2}dy = -4 \int \frac{du}{u} = -4\log|u| + C = -4\log|4-y^2| + C $