$\int\left(2x^4-3\right)^6x^3dx$
$\lim_{x\to\infty}\left(\frac{x-1}{x^2+2x+4}\right)$
$\frac{8x^3-1}{2x-1}$
$\sec^2\left(x\right)-\tan\left(x\right)=1$
$\frac{dy}{dx}=cos8x$
$2\left(4+a\right)$
$\int_0^{0.1}\pi\:\left(17.15x\right)^2dx$
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