Answer
$$A = \frac{1}{{10}}\ln \left( 9 \right)$$
Work Step by Step
$$\eqalign{
& {\text{The area of the region is given by}} \cr
& A = \int_0^4 {\frac{1}{{25 - {x^2}}}} dx \cr
& {\text{Integrate by tables, use }}\int {\frac{1}{{{a^2} - {u^2}}}} = - \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C \cr
& A = - \frac{1}{{2\left( 5 \right)}}\left[ {\ln \left| {\frac{{x - 5}}{{x + 5}}} \right|} \right]_0^4 \cr
& {\text{Evaluate}} \cr
& A = - \frac{1}{{10}}\left[ {\ln \left| {\frac{{4 - 5}}{{4 + 5}}} \right| - \ln \left| {\frac{{0 - 5}}{{0 + 5}}} \right|} \right] \cr
& A = - \frac{1}{{10}}\left[ {\ln \left( {\frac{1}{9}} \right) - \ln \left( 1 \right)} \right] \cr
& A = - \frac{1}{{10}}\ln \left( {\frac{1}{9}} \right) \cr
& A = \frac{1}{{10}}\ln \left( 9 \right) \cr} $$
