Question

$\int\frac{1}{x^{2}-25}dx=$

• A. $\log|\frac{x-5}{x+5}|+C$
• B. $\log|\frac{x+5}{x-5}|+C$
• C. $\frac{1}{5}\log|\frac{x-5}{x+5}|+C$
• D. $\frac{1}{10}\log|\frac{x-5}{x+5}|+C$
• E. $\frac{1}{5}\log|\frac{x+5}{x-5}|+C$

Hint:

Integration Tip: If the denominator is $a^2 - x^2$ instead of $x^2 - a^2$, the formula flips to $\frac{1}{2a} \log|\frac{a+x}{a-x}| + C$. Always pay attention to the order of subtraction!

Answer 1

The Correct Option is D

Concept:
The integral of a rational function of the form $\frac{1}{x^2 - a^2}$ is a standard formula derived using partial fraction decomposition. The standard identity is $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$.

Step 1: Identify the structure of the integrand.
The given integral is: $$I = \int \frac{1}{x^2 - 25} dx$$

Step 2: Express the constant as a perfect square.
Rewrite $25$ as $5^2$ to match the standard identity format $x^2 - a^2$: $$I = \int \frac{1}{x^2 - 5^2} dx$$

Step 3: Determine the value of parameter a.
By comparing the denominator to $x^2 - a^2$, we can clearly see that: $$a = 5$$

Step 4: Apply the standard integration formula.
Substitute $a = 5$ into the identity $\frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$: $$I = \frac{1}{2(5)} \log \left| \frac{x-5}{x+5} \right| + C$$

Step 5: Simplify the resulting expression.
Multiply the constants in the denominator to get the final answer: $$I = \frac{1}{10} \log \left| \frac{x-5}{x+5} \right| + C$$ Hence the correct answer is (D) $\frac{1{10}\log|\frac{x-5}{x+5}|+C$}.