# What is the net area between f(x) = 2/(x+1)^2  and the x-axis over x in [1, 2 ]?

$\frac{1}{3}$
$\frac{2}{x + 1} ^ 2 > 0 , \forall x \ne - 1$
${\int}_{1}^{2} \frac{2}{x + 1} ^ 2 \mathrm{dx} = - 2 {\left[\frac{1}{x + 1}\right]}_{1}^{2} = - 2 \left(\frac{1}{3} - \frac{1}{2}\right) = - 2 \left(- \frac{1}{6}\right) = \frac{1}{3}$