$2\:\ln\left(x\right)-\ln\left(5\right)=\ln\left(x+10\right)$
$\lim_{x\to\infty}x^2\cdot\sin\left(\frac{1}{x^2}\right)$
$\frac{1}{sec\:x}$
$\int\left(\left(3x-5\right)^2\right)dx$
$\frac{12x\:+\:8}{3x^2+4x}$
$-806-928$
$\frac{25n^{12}-625}{5n^6+25}$
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