$\frac{x^2+6x+8}{x+3}$
$\lim_{x\to\infty}\left(\frac{1}{6x+3}\right)$
$\left(-18\right)+\left(-6\right)-\left(-12\right)+\left(-4\right)$
$3a\left(2a+5\right)$
$\lim_{x\to +\infty}\left(3+\frac{2}{x}\right)^{4x}$
$28=9x-8$
$\int\frac{10x-25}{2x^2-11x+12}dx$
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!