$2x+5>=9$
$\frac{x+2}{x}<4$
$3\sin\:2x=\cos\left(x\right)$
$du+5udu=5xdx$
$cos\left(-\frac{1}{7}\right)cos\left(-\frac{1}{4}\right)-sin\left(-\frac{1}{7}\right)sin\left(-\frac{1}{4}\right)$
$\frac{3x^2}{6x^2+3}$
$16m^2-64m^2-64m^2$
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