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