## Calculus: Early Transcendentals (2nd Edition)

$$\ln 2 - \frac{3}{4}$$
\eqalign{ & \int_{ - 1}^1 {\frac{x}{{{{\left( {x + 3} \right)}^2}}}dx} \cr & {\text{integrand }} \cr & = \frac{x}{{{{\left( {x + 3} \right)}^2}}} \cr & {\text{partial fractions}} \cr & \frac{x}{{{{\left( {x + 3} \right)}^2}}} = \frac{A}{{x + 3}} + \frac{B}{{{{\left( {x + 3} \right)}^2}}} \cr & x = A\left( {x + 3} \right) + B \cr & {\text{multiplying}} \cr & x = Ax + 3A + B \cr & x = Ax + \left( {3A + B} \right) \cr & {\text{by equating the coefficients}} \cr & x:{\text{ }}A = 1 \cr & {x^0}:{\text{ }}3A + B = 0 \cr & {\text{Solving these equations}} \cr & A = 1 \cr & B = - 3 \cr & {\text{substituting constants}} \cr & \frac{x}{{{{\left( {x + 3} \right)}^2}}} = \frac{1}{{x + 3}} - \frac{3}{{{{\left( {x + 3} \right)}^2}}} \cr & \int_{ - 1}^1 {\frac{x}{{{{\left( {x + 3} \right)}^2}}}dx} = \int_{ - 1}^1 {\left( {\frac{1}{{x + 3}} - \frac{3}{{{{\left( {x + 3} \right)}^2}}}} \right)dx} \cr & {\text{integrating}} \cr & = \left. {\left( {\ln \left| {x + 3} \right| + \frac{3}{{x + 3}}} \right)} \right|_{ - 1}^1 \cr & {\text{evaluate limits}} \cr & = \left( {\ln \left| {1 + 3} \right| + \frac{3}{{1 + 3}}} \right) - \left( {\ln \left| { - 1 + 3} \right| + \frac{3}{{ - 1 + 3}}} \right) \cr & = \ln 4 + \frac{3}{4} - \ln 2 - \frac{3}{2} \cr & = \ln \left( {\frac{4}{2}} \right) - \frac{3}{4} \cr & = \ln 2 - \frac{3}{4} \cr}