Answer
$$\ln \left| {\frac{{x - 5}}{{x + 5}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{10}}{{{x^2} - 25}}} dx \cr
& or \cr
& = 10\int {\frac{1}{{{x^2} - {{\left( 5 \right)}^2}}}} dx \cr
& {\text{integrate by tables using the formulas on the apendix D for this book}} \cr
& {\text{using the formula 8}}:\,\,\,\,\,\,\,\int {\frac{1}{{{x^2} - {a^2}}}dx = \frac{1}{{2a}} \cdot \ln \left| {\frac{{x - a}}{{x + a}}} \right|} + C{\text{ with }}a \ne 0 \cr
& {\text{in the integral we can see that }}a = 5 \cr
& 10\int {\frac{1}{{{x^2} - {{\left( 5 \right)}^2}}}} dx = 10\left( {\frac{1}{{2\left( 5 \right)}}\ln \left| {\frac{{x - 5}}{{x + 5}}} \right|} \right) + C \cr
& {\text{simplifying}} \cr
& 10\int {\frac{1}{{{x^2} - {{\left( 5 \right)}^2}}}} dx = \ln \left| {\frac{{x - 5}}{{x + 5}}} \right| + C \cr} $$
