Chapter 8: Techniques of Integration - Practice Exercises - Page 517: 31

$$ - \frac{1}{2}\ln \left| {4 - {x^2}} \right| + C$$

$$\eqalign{ & \int {\frac{{xdx}}{{4 - {x^2}}}} \cr & \cr & \left( {\bf{a}} \right){\text{using the substitution method}} \cr & \,\,\,\,{\text{Let }}u = 4 - {x^2},\,\,\,\,\,du = - 2xdx,\,\,\,\,dx = \frac{{du}}{{ - 2x}} \cr & \int {\frac{{xdx}}{{4 - {x^2}}}} = \int {\frac{x}{u}\left( {\frac{{du}}{{ - 2x}}} \right)} \cr & = - \frac{1}{2}\int {\frac{1}{u}du} \cr & {\text{integrate }} \cr & = - \frac{1}{2}\ln \left| u \right| + C \cr & {\text{write in terms of }}x,{\text{ substitute }}4 - {x^2}{\text{ for }}u \cr & = - \frac{1}{2}\ln \left| {4 - {x^2}} \right| + C \cr & \cr & \left( {\bf{b}} \right){\text{using a trigonometric substitution}} \cr & \,\,\,\,\,{\text{Let }}x = 2\sin \theta ,\,\,\,dx = 2\cos \theta d\theta \cr & \,\,\,\,\,\int {\frac{{xdx}}{{4 - {x^2}}}} = \int {\frac{{\left( {2\sin \theta } \right)\left( {2\cos \theta d\theta } \right)}}{{4 - {{\left( {2\sin \theta } \right)}^2}}}} \cr & = \int {\frac{{4\sin \theta \cos \theta d\theta }}{{4 - 4{{\sin }^2}\theta }}} \cr & = \int {\frac{{\sin \theta \cos \theta d\theta }}{{1 - {{\sin }^2}\theta }}} \cr & = \int {\frac{{\sin \theta \cos \theta d\theta }}{{{{\cos }^2}\theta }}} \cr & = \int {\frac{{\sin \theta d\theta }}{{\cos \theta }}} \cr & = \int {\tan \theta } d\theta \cr & {\text{Integrate}} \cr & = - \ln \left| {\cos \theta } \right| + C \cr & {\text{where }}\cos \theta = \frac{{\sqrt {4 - {x^2}} }}{2} \cr & = - \ln \left| {\frac{{\sqrt {4 - {x^2}} }}{2}} \right| + C \cr & {\text{simplify using logaritmic properties and combine the constants}} \cr & = - \ln \left| {\sqrt {4 - {x^2}} } \right| + \ln \left| 2 \right| + C \cr & = - \frac{1}{2}\ln \left| {4 - {x^2}} \right| + C \cr} $$