# 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?

*print*Print*list*Cite

The area of the region enclosed by the curve y = 1/x + 1/(x+1), the lines x=1 and x=2 and the x-axis is the definite integral of y = 1/x + 1/(x + 1) between the limits x = 1 and x = 2.

Int[ y dx], x = 1 to x = 2

=> Int [ 1/x + 1/(x + 1) dx], x = 1 to x = 2

=> ln|x| + ln|x + 1|, x = 1 to x = 2

=> ln 2 + ln 3 - ln 1 - ln 2

=> ln 3 - ln 1

=> ln 3

**The required area is ln 3**

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.**