$6x+8<3x-8$
$\lim_{x\to\infty}37x+2$
$25m^4-40m^2+16$
$\int\left(\frac{6x^3-5x}{3x^4-5x^2}\right)dx$
$10y-2y=64$
$\sec\left(a\right)-\cos\left(a\right)=\tan\left(a\right)\cdot\sin\left(a\right)$
$\int x^2\tanh\left(x^3\right)dx$
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