$\int_{\infty}^{-1}\left(\frac{1}{x^{\frac{1}{3}}}\right)dx$

$\int e^{-0.06t}dx$

$\int\frac{x}{x^4-4x^2+3}dx$

$f\left(x\right)=\left(x-1\right)\left(x^2-3x+2\right)$

$2-12\pi$

$\sin\left(x\right)^2-4\sin\left(x\right)-5=0$

$\frac{dy}{dx}-\left(\sin\left(x\right)\right)y=2\sin\left(x\right)$