$\frac{\sin\left(7x\right)+sin\left(x\right)}{sin\left(4x\right)}=1$

$\int_{-\infty}^{-3}\left(\frac{1}{x^2+4x+3}\right)dx$

$\frac{x\left(x+4\right)}{x\left(x+4\right)+x\left(x+5\right)+4}$

$9x^2-94x+176$

$\frac{dy}{dx}=\frac{3x}{5y+1}$

$-4b\left(5b+3\right)+2\left(2b-3\right)$

$\frac{-3x^3}{3x}$