Answer
$${\text{the improper integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^5 {\frac{1}{{25 - {x^2}}}} dx \cr
& \frac{1}{{25 - {x^2}}}{\text{ has an infinite discontinuity at }}x = 5,{\text{so we can write}} \cr
& \int_0^5 {\frac{1}{{25 - {x^2}}}} dx = \mathop {\lim }\limits_{b \to {5^ - }} \int_0^b {\frac{1}{{25 - {x^2}}}} dx \cr
& {\text{Integrate by tables, use }}\int {\frac{1}{{{a^2} - {x^2}}}} dx = \frac{1}{{2a}}\ln \left| {\frac{{x + a}}{{x - a}}} \right| + C \cr
& \mathop {\lim }\limits_{b \to {5^ - }} \int_0^b {\frac{1}{{25 - {x^2}}}} dx = \frac{1}{{2\left( 5 \right)}}\mathop {\lim }\limits_{b \to {5^ - }} \left[ {\ln \left| {\frac{{x + 5}}{{x - 5}}} \right|} \right]_0^b \cr
& = \frac{1}{{10}}\mathop {\lim }\limits_{b \to {5^ - }} \left[ {\ln \left| {\frac{{x + 5}}{{x - 5}}} \right|} \right]_0^b \cr
& = \frac{1}{{10}}\mathop {\lim }\limits_{b \to {5^ - }} \left[ {\ln \left| {\frac{{b + 5}}{{b - 5}}} \right| - \ln \left| {\frac{{0 + 5}}{{0 - 5}}} \right|} \right] \cr
& = \frac{1}{{10}}\mathop {\lim }\limits_{b \to {5^ - }} \left[ {\ln \left| {\frac{{b + 5}}{{b - 5}}} \right|} \right] \cr
& {\text{Evaluate the limit when }}b \to {5^ - } \cr
& = \frac{1}{{10}}\left[ {\ln \left| {\frac{{5 + 5}}{{0 - 5}}} \right|} \right] \cr
& = \frac{1}{{10}}\left( \infty \right) \cr
& = \infty \cr
& {\text{So}},{\text{ the improper integral diverges}}. \cr
& {\text{The following graph confirms the result}} \cr} $$
