What is the area of the region under the curve y=1/x + 1/(x+1), beween the lines x=1 and x=2 and x axis?

In other words, we'll have to determine the area of the
region bounded by the given curve, x axis and the lines x = 1 and x 
=2.

We'll evaluate the definite integral of the function y
between the limits of integration: x = 1 to x = 2.

Int ydx
= Int [1/x + 1/(x+1)]dx

We'll apply the additive property
of integrals:

Int [1/x + 1/(x+1)]dx = Int dx/x + Int
dx/(x+1)

Int dx/x + Int dx/(x+1) = ln |x| + ln|x +
1|

We'll apply Leibniz Newton formula to evaluate the
definite integral:

Int dx/x + Int dx/(x+1) = F(2) -
F(1)

Int dx/x + Int dx/(x+1) = ln 2 + ln 3 - ln 1 - ln
2

But ln 1 = 0

We'll reduce
like terms and we'll get:

Int dx/x + Int dx/(x+1) = ln
3

The area of the region bounded by the given
curve, x axis and the lines x = 1 and x  =2 is A = ln 3 square
units.