$\int\frac{.\left(2x+3\right)}{x\left(x+4\right)^2}dx$
$\frac{8x^4-3^3-7x^2-6}{2x^2-5x+6}$
$\frac{64x^6-b^{12}}{2x+b^2}$
$\int7x^9dx$
$4x^2+6x=12$
$\int\pi\left(9-8\sqrt{-x^2+8x-7\:}-x^2+8x\right)dx$
$7x^2+4x+7$
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