Answer
$$6\sqrt 3 $$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^2 {\frac{3}{{{{\left( {x + 1} \right)}^{1/2}}}}} dx \cr
& {\text{The integrand }}\frac{3}{{{{\left( {x + 1} \right)}^{1/2}}}}{\text{ is not continuous at }}x = - 1,{\text{ then by }} \cr
& {\text{the definition of improper integrals}} \cr
& \int_{ - 1}^2 {\frac{3}{{{{\left( {x + 1} \right)}^{1/2}}}}} dx = \mathop {\lim }\limits_{r \to - {1^ + }} \int_r^2 {\frac{3}{{{{\left( {x + 1} \right)}^{1/2}}}}} dx \cr
& {\text{Integrating }} \cr
& = \mathop {\lim }\limits_{r \to - {1^ + }} \left[ {\frac{{3{{\left( {x + 1} \right)}^{1/2}}}}{{1/2}}} \right]_r^2 \cr
& = 6\mathop {\lim }\limits_{r \to - {1^ + }} \left[ {{{\left( {x + 1} \right)}^{1/2}}} \right]_r^2 \cr
& = 6\mathop {\lim }\limits_{r \to - {1^ + }} \left[ {{{\left( {2 + 1} \right)}^{1/2}} - {{\left( {r + 1} \right)}^{1/2}}} \right] \cr
& = 6\mathop {\lim }\limits_{r \to - {1^ + }} \left[ {\sqrt 3 - {{\left( {r + 1} \right)}^{1/2}}} \right] \cr
& {\text{Evaluate the limit when }}r \to - {1^ + } \cr
& = 6\left[ {\sqrt 3 - {{\left( { - {1^ + } + 1} \right)}^{1/2}}} \right] \cr
& = 6\left[ {\sqrt 3 } \right] \cr
& = 6\sqrt 3 \cr} $$
