$\int\:\frac{sec^3\left(\sqrt{x}\right)tan\left(\sqrt{x}\right)}{\sqrt{x}}dx$
$\int\left(2x\left(x^2+6\right)^4\right)dx$
$\left(7b-2x\right)^2$
$-4x^3y-4xy^2;\:x=2;\:y=-1$
$20x=20x+200$
$\frac{26-36x^4}{5-6x^2}$
$\frac{\sqrt{168x^4}}{\sqrt{6x}}$
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