$\int\:\frac{\left(x+1\right)}{x^2+4x-5}dx$
$3cos^2x-1$
$3x^2-4x^4$
$-6\left(-6c+6\right)-5$
$x-y^2=0$
$\left(25+56+89\right)+59$
$\frac{d}{dx}\left(x^2+-3\ln\left(y\right)+y^2=16\right)$
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