Question

What is the net area between `f(x) = 1/sqrt(x^2+2x+1)` and the x-axis over `x in [2, 4 ]`?

Answer 1

The answer is ln(5/3)

Explanation:

`sqrt(x^2+2x+1)=abs(1+x)` so `f(x)=1/abs(1+x)` the area is the `int_2^4f(x)dx=ln(1+x)|_2^4=ln(5)-ln(3)=ln(5/3)`