$\int\frac{18-5x}{\left(x-3\right)^2}dx$

$\:5a+10b=-2b-4a\:$

$\int\frac{x^2-3}{\left(x+1\right)^3}dx$

$\int\frac{senx}{cosx+cos^2x}dx$

$\frac{dy}{dx}=x^2+3x+1$

$\frac{\sin\left(x\right)}{1-\cos\left(x\right)}+\frac{\sin\left(x\right)}{1+\cos\left(x\right)}=2\sin\left(x\right)$

$\left(10a+3b\right)\cdot\left(10a-3b\right)$