$\int\frac{6x^2+1}{x^3-3x^2}dx$

$\left(2x^3-3x^2-4x-5\right)\frac{4}{3}$

$\sin\left(x\right)\left(-\sin\left(x\right)\right)-\left(\cos\left(x\right)\left(\cos\left(x\right)\right)\right)$

$x^2+6x-11$

$36x^{4m}-9y^6$

$4\sin\left(x\right)-1=0$

$\cot\theta=\cos\theta\sin\theta+\cos^{3}\theta\csc\theta$