$tan^{-1}\left(\frac{2}{3}\right)-\:tan^{-1}\left(\frac{x}{5}\right)=\:\frac{\pi}{4}$
$\frac{x+1}{\left(x-4\right)}\cdot\frac{1}{\left(x+2\right)}$
$\int_0^{2\pi}4sen^2xcos^3xdx$
$2x^2+9x=-4$
$\frac{8x^2-2x}{2x}$
$y'-\left(5y^2\right)+y=0;y\left(0\right)=0$
$-\:1\:+\:1\:+\:1\:-\:1$
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