Question

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

Answer 1

`1/3`
graph{(y-2/((x+1)^2))=0 [-0.168, 4.832, -0.2, 2.41]}

Explanation:

`2/(x+1)^2>0, AAx!=-1`

`int_1^2 2/(x+1)^2dx=-2[1/(x+1)]_1^2=-2(1/3-1/2)=-2(-1/6)=1/3`