$\frac{\left(6x^3+3x^2-10x-12\right)}{2x-3}$

$\int3x\left(1+x\right)^{-4}dx$

$\int_0^{\frac{\pi}{4}}\left(\pi tan^3\left(x\right)\right)\left(-\pi sec^3\left(x\right)\right)dx$

$y\sin^2x\frac{dy}{dx}=\cos x$

$-\frac{385}{-55}$

$\frac{15}{3}xy-\frac{2}{6}yx$