Question

The value of \( \int_{2}^{3} \frac{x}{x^2 - 1} \, dx \) is

• A. \( \frac{-1}{2} \log \left( \frac{8}{3} \right) \)
• B. \( \frac{1}{2} \log \left( \frac{8}{3} \right) \)
• C. \( \frac{-1}{3} \log \left( \frac{8}{3} \right) \)
• D. \( \frac{1}{3} \log \left( \frac{8}{3} \right) \)

Hint:

For integrals involving \( \frac{x}{x^2 - 1} \), use substitution \( u = x^2 - 1 \) to simplify the integral.

Answer 1

The Correct Option is B

Step 1: Understanding the integral.
The given integral is of the form \( \int \frac{x}{x^2 - 1} \, dx \), which can be solved using substitution. Let’s use the substitution \( u = x^2 - 1 \), so \( du = 2x \, dx \).

Step 2: Substituting in the integral.
We rewrite the integral: \[ \int \frac{x}{x^2 - 1} \, dx = \frac{1}{2} \int \frac{du}{u} \] The integral of \( \frac{1}{u} \) is \( \ln |u| \), so we have: \[ \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x^2 - 1| + C \]
Step 3: Applying the limits.
Now, we apply the limits of integration \( x = 2 \) to \( x = 3 \): \[ \frac{1}{2} \left( \ln |9 - 1| - \ln |4 - 1| \right) = \frac{1}{2} \left( \ln 8 - \ln 3 \right) = \frac{1}{2} \log \left( \frac{8}{3} \right) \]
Step 4: Conclusion.
Thus, the value of the integral is \( \frac{1}{2} \log \left( \frac{8}{3} \right) \), which makes option (B) the correct answer.