Answer
$$ - \frac{1}{{48}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^{ - 1} {\frac{x}{{{{\left( {{x^2} + 2} \right)}^3}}}dx} \cr
& u = {x^2} + 2,{\text{ then }}du = 2xdx,{\text{ }}\frac{1}{2}du = xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = - 1,{\text{ }}u = {\left( { - 1} \right)^2} + 2 = 3 \cr
& {\text{if }}x = - 2,{\text{ }}u = {\left( { - 2} \right)^2} + 2 = 6 \cr
& {\text{so}} \cr
& \int_{ - 2}^{ - 1} {\frac{x}{{{{\left( {{x^2} + 2} \right)}^3}}}dx} = \int_6^3 {\frac{1}{{{u^3}}}\left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int_6^3 {{u^{ - 3}}du} \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{ - 2}}}}{{ - 2}}} \right)_6^3 \cr
& = - \frac{1}{4}\left( {\frac{1}{{{u^4}}}} \right)_6^3 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \frac{1}{4}\left( {\frac{1}{{{{\left( 3 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}} \right) \cr
& = - \frac{1}{4}\left( {\frac{1}{9} - \frac{1}{{36}}} \right) \cr
& = - \frac{1}{4}\left( {\frac{1}{{12}}} \right) \cr
& = - \frac{1}{{48}} \cr} $$
