$\int\left(\left(\frac{\left(x+1\right)}{\left(4x-3\right)^2}\right)\right)dx$

$\sin\left(\frac{\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)$

$-\frac{e^{-0.12\left(0\right)}}{-0.12}$

$\frac{d}{dx}\:y=\sqrt[4]{3x-5}$

$\int\left(\frac{1}{\left(1-3t\right)^4}\right)dt$

$65.47\cdot106.7$

$\int\frac{-2}{x}dx$