Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 542: 64

$2\ln 2-\ln 3$

The area is: $$\int_1^2 \frac{1}{x^2+x}dx$$ We have: $$\int_1^2 \frac{1}{x^2+x}dx=\int_1^2 \frac{1}{x(x+1)}dx=\int_1^2 \left(\frac{1}{x}-\frac{1}{x+1}\right)dx=[\ln x-\ln(x+1)]_1^2 =\left[\ln\left(\frac{x}{x+1}\right)\right]_1^2=\ln\frac{2}{3}-\ln\frac{1}{2}=\ln\frac{4}{3}=2\ln 2-\ln 3$$