Answer
$$ - \sqrt 8 $$
Work Step by Step
$$\eqalign{
& \int_{ - 3}^1 {\frac{{xdx}}{{\sqrt {9 - {x^2}} }}} \cr
& {\text{The function is undefined for }}x = - 3,{\text{ so the integral can be repressented as}} \cr
& \int_{ - 3}^1 {\frac{{xdx}}{{\sqrt {9 - {x^2}} }}} = \mathop {\lim }\limits_{k \to {0^ + }} \int_k^1 {\frac{{xdx}}{{\sqrt {9 - {x^2}} }}} \cr
& {\text{Integrate}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \mathop {\lim }\limits_{k \to {{\left( { - 3} \right)}^ + }} \left[ {\sqrt {9 - {x^2}} } \right]_k^1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \mathop {\lim }\limits_{k \to {{\left( { - 3} \right)}^ + }} \left[ {\sqrt {9 - {{\left( 1 \right)}^2}} - \sqrt {9 - {k^2}} } \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \mathop {\lim }\limits_{k \to {{\left( { - 3} \right)}^ + }} \left[ {\sqrt 8 - \sqrt {9 - {k^2}} } \right] \cr
& \,{\text{Calculate the limit when }}k \to {0^ + } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \left[ {\sqrt 8 - \sqrt {9 - {{\left( { - 3} \right)}^2}} } \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \sqrt 8 \cr
& {\text{Then}}{\text{,}} \cr
& \int_{ - 3}^1 {\frac{{xdx}}{{\sqrt {9 - {x^2}} }}} = - \sqrt 8 \cr} $$
