Answer
$$\frac{1}{5}\ln \left( 3 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{1}{{25 - {x^2}}}} dx \cr
& = \int_0^4 {\frac{1}{{{{\left( 5 \right)}^2} - {x^2}}}} dx \cr
& {\text{From Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{{a^2} - {u^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{a + u}}{{a - u}}} \right| + C,{\text{ then}} \cr
& \int_0^4 {\frac{1}{{{{\left( 5 \right)}^2} - {x^2}}}} dx = \left[ {\frac{1}{{2\left( 5 \right)}}\ln \left| {\frac{{5 + x}}{{5 - x}}} \right|} \right]_0^4 \cr
& = \frac{1}{{10}}\left[ {\ln \left| {\frac{{5 + x}}{{5 - x}}} \right|} \right]_0^4 \cr
& {\text{Evaluating}} \cr
& = \frac{1}{{10}}\left[ {\ln \left| {\frac{{5 + 4}}{{5 - 4}}} \right| - \ln \left| {\frac{{5 + 0}}{{5 - 0}}} \right|} \right] \cr
& = \frac{1}{{10}}\left[ {\ln \left| 9 \right| - \ln \left| 1 \right|} \right] \cr
& = \frac{1}{{10}}\ln \left( 9 \right) \cr
& {\text{Simplifying}} \cr
& = \frac{1}{{10}}\ln \left( {{3^2}} \right) \cr
& = \frac{1}{5}\ln \left( 3 \right) \cr} $$
