Step 1: Understanding the Concept:
This integral is of a standard form that can be solved either by using partial fraction decomposition or by applying a standard integration formula.
Step 2: Key Formula or Approach:
The standard integration formula for this type of integrand is:
\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left|\frac{a+x}{a-x}\right| + C \]
Alternatively, using partial fractions:
\[ \frac{1}{a^2-x^2} = \frac{1}{(a-x)(a+x)} = \frac{A}{a-x} + \frac{B}{a+x} \]
Step 3: Detailed Explanation:
The given integral is $\int \frac{dx}{25 - x^2}$.
This can be written as $\int \frac{dx}{5^2 - x^2}$.
This matches the standard form $\int \frac{dx}{a^2 - x^2}$ with $a=5$.
Applying the formula:
\[ \int \frac{dx}{5^2 - x^2} = \frac{1}{2(5)} \log \left|\frac{5+x}{5-x}\right| + C \]
\[ = \frac{1}{10} \log \left|\frac{5+x}{5-x}\right| + C \]
Now let's examine the options. Option (C) is $\frac{1}{10} \log \left|\frac{5+x}{x-5}\right| + c$.
We know that $5-x = -(x-5)$.
Therefore, $\left|\frac{5+x}{5-x}\right| = \left|\frac{5+x}{-(x-5)}\right| = \frac{|5+x|}{|-(x-5)|} = \frac{|5+x|}{|x-5|} = \left|\frac{5+x}{x-5}\right|$.
So, our result $\frac{1}{10} \log \left|\frac{5+x}{5-x}\right| + C$ is equivalent to option (C).
Step 4: Final Answer:
The value of the integral is $\frac{1}{10} \log \left|\frac{5+x}{5-x}\right| + C$, which is equivalent to the expression in option (C). Therefore, option (C) is correct.