$\frac{dy}{dx}=1+\sqrt{y-2x+3}$
$\int_1^{\infty}\left(\frac{x+4}{x^2}\right)dx$
$\left(-2^4\right)+\left(3^2\right)-\left(-4^2\right)+\left(-1^3\right)-\left(5^2\right)$
$x-2+x-2$
$\frac{3+x}{5+x}=\frac{4}{5}$
$\int\frac{8}{x^2-4}dx$
$x^2+40x+150$
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