$2x\tan\left(y\right)dx+\sec\left(y\right)^2dy=0$
$x^3+2xy-4=0$
$\int\left(\frac{x^3+x^2-3x-4}{x^2+4x+3}\right)dx$
$\int\tan\left(-x\right).\sec\left(-x\right)^2dx$
$\frac{x+4}{3}+\frac{x-2}{2}$
$\frac{8x^3+16x+6x}{2x+3}$
$\frac{dx}{dy}+\frac{yx}{y^2-9}=0$
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