$-1+\:-1$
$\int\tan^5\left(2x+5\right)dx$
$\left(x\right)\frac{dy}{dx}=\left(1-4x^2\right)\tan\left(y\right)$
$\lim_{x\to2}\left(\frac{x^2-4}{x^2-6x+8}\right)$
$\frac{-1}{30}x^2+\frac{5}{4}x\:-\:\sqrt{5}$
$\int\frac{3x}{\sqrt[5]{\left(2-x\right)^4}}dx$
$5^1$
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