solve the following definite integral by decomposition into partial fractions: \int_{1}^{2}{\frac{1}{x^2+x}}dx

\frac{1}{x^2+x}=\frac{A}{x} + \frac{B}{x+1}, 1=A(x+1)+Bx, let x=-1: 1=-B, B=-1let x=0: 1=A, A=1Hence, \int_{1}^{2}{\frac{1}{x^2+x}dx} = \int_{1}^{2}{\frac{1}{x}dx} - \int_{1}^{2}{\frac{1}{x+1}dx}=[ln(x)-ln(x+1)]{1}^{2}=[ln(/frac{x}{x+1})]{1}^{2}=ln(\frac{2}{3})-ln(\frac{1}{2})=ln(\frac{4}{3})=2*ln(2)-ln(3)