Answer
$\frac{5}{2}\ln \left| {\frac{{x - 5}}{{x + 5}}} \right| + C$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{25}}{{{x^2} - 25}} \cr
& {\text{Separate the variables}} \cr
& dy = \frac{{25}}{{{x^2} - 25}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{{25}}{{{x^2} - 25}}} dx \cr
& y = 25\int {\frac{1}{{{x^2} - 25}}} dx \cr
& {\text{Integrate by tables}}{\text{, using the formula}} \cr
& {\text{* }}\int {\frac{1}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right|} + C \cr
& {\text{Therefore}}{\text{,}} \cr
& y = 25\left( {\frac{1}{{2\left( 5 \right)}}\ln \left| {\frac{{x - 5}}{{x + 5}}} \right|} \right) + C \cr
& y = \frac{5}{2}\ln \left| {\frac{{x - 5}}{{x + 5}}} \right| + C \cr} $$
