$$\displaystyle\int{ \frac{ 1 }{ \sqrt{ 16- { \left(3x+4 \right) }^{ 2 } } } }d x$$
$\frac{\arcsin(\frac{3x+4}{4})}{3}+С$
$\frac{\sqrt{\frac{3}{x\left(-3x-8\right)}}}{3}$
