$\left(4+x\right)^2-\frac{16}{\left(4+x\right)+4}$
$6xe^{-2x}-\int\left(6e^{-2x}\right)du$
$4\sin^2\left(x\right)\cos^2\left(x\right)$
$\frac{10x^2y^5}{2y^2}\:x\:\frac{4\left(x^2\right)^5}{y^{10}}$
$x^4-2x^3+x^2+x-1$
$\int\frac{1}{\sqrt{6r-1}}dr$
$\int\frac{\left(\sqrt{4x^2-1}\right)}{5x}dx$
Get a preview of step-by-step solutions.
Earn solution credits, which you can redeem for complete step-by-step solutions.
Save your favorite problems.
Become premium to access unlimited solutions, download solutions, discounts and more!