Question

\(\int \frac{dx}{25 - x^2} =\)

• A. \(\frac{1}{5} \log \left| \frac{x-5}{x+5} \right| + c\)
• B. \(\frac{1}{5} \log \left| \frac{x+5}{x-5} \right| + c\)
• C. \(\frac{1}{10} \log \left| \frac{5+x}{5-x} \right| + c\)
• D. \(\frac{1}{10} \log \left| \frac{5-x}{5+x} \right| + c\)

Hint:

Memorizing the standard integral forms for \(\frac{1}{a^2-x^2}\), \(\frac{1}{x^2-a^2}\), and \(\frac{1}{a^2+x^2}\) is essential for speed and accuracy in competitive exams. They appear very frequently.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to evaluate the indefinite integral of \(\frac{1}{25 - x^2}\).

Step 2: Key Formula or Approach:
This integral matches the standard integration formula:
\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \]
Alternatively, we can use partial fraction decomposition.

Step 3: Detailed Explanation:

Method 1: Using the Standard Formula
The integral is \(\int \frac{dx}{25 - x^2}\).
We can write this as \(\int \frac{dx}{5^2 - x^2}\).
This matches the standard form with \(a=5\).
Applying the formula:
\[ \int \frac{dx}{5^2 - x^2} = \frac{1}{2(5)} \ln \left| \frac{5+x}{5-x} \right| + c \]
\[ = \frac{1}{10} \ln \left| \frac{5+x}{5-x} \right| + c \]
Assuming \(\log\) in the options means natural logarithm (\(\ln\)), this matches option (C).

Method 2: Partial Fraction Decomposition
Factor the denominator: \(25 - x^2 = (5-x)(5+x)\).
Decompose the fraction:
\[ \frac{1}{(5-x)(5+x)} = \frac{A}{5-x} + \frac{B}{5+x} \]
Using the cover-up method:
To find A, cover \((5-x)\) and set \(x=5\): \(A = \frac{1}{5+5} = \frac{1}{10}\).
To find B, cover \((5+x)\) and set \(x=-5\): \(B = \frac{1}{5-(-5)} = \frac{1}{10}\).
So the integral becomes:
\[ \int \left( \frac{1/10}{5-x} + \frac{1/10}{5+x} \right) dx = \frac{1}{10} \int \frac{dx}{5-x} + \frac{1}{10} \int \frac{dx}{5+x} \]
\[ = \frac{1}{10} (-\ln|5-x|) + \frac{1}{10} (\ln|5+x|) + c \]
\[ = \frac{1}{10} (\ln|5+x| - \ln|5-x|) + c \]
Using the logarithm property \(\ln a - \ln b = \ln(a/b)\):
\[ = \frac{1}{10} \ln \left| \frac{5+x}{5-x} \right| + c \]

Step 4: Final Answer:
The result of the integration is \(\frac{1}{10} \log \left| \frac{5+x}{5-x} \right| + c\).