Answer
$$ - \frac{{2\sqrt {4 - {w^2}} }}{w} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{8dw}}{{{w^2}\sqrt {4 - {w^2}} }}} \cr
& = 8\int {\frac{{dw}}{{{w^2}\sqrt {{2^2} - {w^2}} }}} \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}w = 2\sin \theta,{\text{ }}dw = 2\cos \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& 8\int {\frac{{dw}}{{{w^2}\sqrt {{2^2} - {w^2}} }}} = 8\int {\frac{{2\cos \theta d\theta }}{{4{{\sin }^2}\theta \sqrt {4 - 4{{\sin }^2}\theta } }}} \cr
& = 4\int {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta \sqrt {4\left( {1 - {{\sin }^2}\theta } \right)} }}} \cr
& {\text{use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& = 2\int {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta \sqrt {{{\cos }^2}\theta } }}} \cr
& {\text{simplifying}} \cr
& = 2\int {\frac{{d\theta }}{{{{\sin }^2}\theta }}} \cr
& = 2\int {{{\csc }^2}\theta } d\theta \cr
& {\text{Integrate}} \cr
& = 2\left( { - \cot \theta } \right) + C \cr
& = - 2\cot \theta + C \cr
& {\text{write in terms of }}x,{\text{ }}\cot \theta = \frac{{\sqrt {4 - {w^2}} }}{w} \cr
& = - 2\left( {\frac{{\sqrt {4 - {w^2}} }}{w}} \right) + C \cr
& = - \frac{{2\sqrt {4 - {w^2}} }}{w} + C \cr} $$
