Answer
$$A = \frac{1}{{81}} + \frac{{\ln 3}}{{108}}$$
Work Step by Step
$$\eqalign{
& {\text{The area under de curve is given by}} \cr
& A = \int_0^{3/2} {{{\left( {9 - {x^2}} \right)}^{ - 2}}} dx \cr
& A = \int_0^{3/2} {\frac{{dx}}{{{{\left( {9 - {x^2}} \right)}^2}}}} \cr
& {\text{The integrand contains the form }}{a^2} - {x^2} \cr
& 9 - {x^2} \to a = 3 \cr
& {\text{Use the change of variable }}x = a\sin \theta \cr
& x = 3\sin \theta ,\,\,\,dx = 3\cos \theta d\theta \cr
& = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {9 - 9{{\sin }^2}\theta } \right)}^2}}}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{81{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}} \cr
& = \frac{1}{{27}}\int {\frac{{\cos \theta d\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^2}}}} \cr
& = \frac{1}{{27}}\int {\frac{{\cos \theta d\theta }}{{{{\cos }^4}\theta }}} \cr
& = \frac{1}{{27}}\int {{{\sec }^3}\theta } d\theta \cr
& {\text{Integrating}} \cr
& = \frac{1}{{27}}\left[ {\frac{1}{2}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right)} \right] + C \cr
& = \frac{1}{{54}}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{{54}}\left( {\left( {\frac{3}{{\sqrt {9 - {x^2}} }}} \right)\left( {\frac{x}{{\sqrt {9 - {x^2}} }}} \right) + \ln \left| {\frac{3}{{\sqrt {9 - {x^2}} }} + \frac{x}{{\sqrt {9 - {x^2}} }}} \right|} \right) + C \cr
& = \frac{x}{{18\left( {9 - {x^2}} \right)}} + \frac{1}{{54}}\ln \left| {\frac{{3 + x}}{{\sqrt {9 - {x^2}} }}} \right| + C \cr
& \cr
& ,{\text{then}} \cr
& A = \int_0^{3/2} {\frac{{dx}}{{{{\left( {9 - {x^2}} \right)}^2}}}} \cr
& A = \left[ {\frac{x}{{18\left( {9 - {x^2}} \right)}} + \frac{1}{{54}}\ln \left| {\frac{{3 + x}}{{\sqrt {9 - {x^2}} }}} \right|} \right]_0^{3/2} \cr
& A = \left[ {\frac{{3/2}}{{18\left( {9 - {{\left( {3/2} \right)}^2}} \right)}} + \frac{1}{{54}}\ln \left| {\frac{{3 + 3/2}}{{\sqrt {9 - 3/{2^2}} }}} \right|} \right]\, - \left[ {\frac{0}{{18\left( {9 - 0} \right)}} + \ln \left| 1 \right|} \right] \cr
& A = \left[ {\frac{1}{{81}} + \frac{1}{{54}}\ln \left| {\sqrt 3 } \right|} \right]\, - \left[ 0 \right] \cr
& A = \frac{1}{{81}} + \frac{1}{{54}}\left( {\frac{1}{2}\ln 3} \right) \cr
& A = \frac{1}{{81}} + \frac{{\ln 3}}{{108}} \cr} $$
