$\int\frac{-2x^2-4x}{\left(x-1\right)^2\left(x^2+1\right)}dx$
$\int\left(\frac{\left(x^2-13x+1\right)}{x^2+x}\right)dx$
$\tan^2x\left(1+\cot^2x\right)$
$\frac{x}{-8}+9>13$
$x^2-32x+256=0$
$y'=3y-4x$
$\frac{dy}{dx}=\:\sqrt{2x\:+y\:-3}-2$
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!