Answer
$${\text{area}} = \frac{1}{{40}}\ln 9$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \frac{1}{{25 - 16{x^2}}},\,\,\,\,y = 0,\,\,\,x = 0,\,\,\,x = 1 \cr
& \cr
& {\text{We can represents the enclosed by the curves as}} \cr
& {\text{area}} = \int_0^1 {\frac{1}{{25 - 16{x^2}}}} dx \cr
& {\text{integrate by tables using the formula }} \cr
& \int {\frac{{du}}{{{a^2} - {u^2}}}} du = \frac{1}{{2a}}\ln \left| {\frac{{u + a}}{{u - a}}} \right| + C \cr
& \cr
& {\text{write }}\int_0^1 {\frac{1}{{25 - 16{x^2}}}} dx{\text{ in terms of }}u \cr
& \,\,\,\,\,\,u = 4x,\,\,\,du = 4dx,\,\, \cr
& \,\,\,\,\,\,x = 0 \to \,\,u = 0{\text{ and }}x = 1 \to \,\,\,u = 4 \cr
& \int_0^1 {\frac{1}{{25 - 16{x^2}}}} dx = \int_0^4 {\frac{1}{{{5^2} - {u^2}}}} \left( {\frac{1}{4}du} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\int_0^4 {\frac{1}{{25 - {u^2}}}} du \cr
& \cr
& {\text{area}} = \frac{1}{4}\int_0^4 {\frac{1}{{25 - {u^2}}}} du \cr
& {\text{area}} = \frac{1}{4}\left[ {\frac{1}{{2\left( 5 \right)}}\ln \left| {\frac{{u + 5}}{{u - 5}}} \right|} \right]_0^4 \cr
& {\text{evaluate the limits and simplify}} \cr
& {\text{area}} = \frac{1}{4}\left[ {\frac{1}{{10}}\ln \left| {\frac{{4 + 5}}{{4 - 5}}} \right| - \frac{1}{{10}}\ln \left| {\frac{{0 + 5}}{{0 - 5}}} \right|} \right] \cr
& {\text{area}} = \frac{1}{{40}}\left[ {\ln \left| {\frac{9}{{ - 1}}} \right| - \ln \left| {\frac{5}{{ - 5}}} \right|} \right] \cr
& {\text{area}} = \frac{1}{{40}}\left( {\ln 9 - \ln 1} \right) \cr
& {\text{area}} = \frac{1}{{40}}\ln 9 \cr} $$
