$x^2+4x+4=9$
$\frac{dy}{dx}=\frac{x^2\ln\left(x\right)-xy}{x^2}$
$\int ln\left(x-6\right)dr$
$\frac{3x+3y}{x}\cdot\frac{1}{3}$
$3m\:^2+\:2mn\:+\:5mn\:+\:6n\:+\:7m^2$
$3=15-x$
$\left(x+4y\right)dx+2xdy=0$
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