## Calculus: Early Transcendentals (2nd Edition)

$$\frac{5}{x} + 6\ln \left| {\frac{{x}}{x+1}} \right| + C$$
\eqalign{ & \int {\frac{{x - 5}}{{{x^2}\left( {x + 1} \right)}}dx} \cr & {\text{partial fractions}} \cr & \frac{{x - 5}}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} \cr & x - 5 = Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C{x^2} \cr & {\text{multiplying}} \cr & x - 5 = A{x^2} + Ax + Bx + B + C{x^2} \cr & x - 5 = \left( {A{x^2} + C{x^2}} \right) + \left( {Ax + Bx} \right) + B \cr & {\text{by equating the coefficients}} \cr & {x^2}:{\text{ }}A + C = 0 \cr & x:{\text{ }}A + B = 1 \cr & {x^0}:{\text{ }}B = - 5 \cr & {\text{Solving these equations}} \cr & A = 6 \cr & B = - 5 \cr & C = - 6 \cr & {\text{substituting constants}} \cr & \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} = \frac{6}{x} - \frac{5}{{{x^2}}} - \frac{6}{{x + 1}} \cr & \int {\frac{{x - 5}}{{{x^2}\left( {x + 1} \right)}}dx} = \int {\left( {\frac{6}{x} - \frac{5}{{{x^2}}} - \frac{6}{{x + 1}}} \right)dx} \cr & {\text{integrating}} \cr & = 6\ln \left| x \right| + \frac{5}{x} - 6\ln \left| {x + 1} \right| + C \cr & {\text{Simplify}} \cr & = \frac{5}{x} + 6\ln \left| {\frac{{x}}{x+1}} \right| + C \cr}