Question

Evaluate \(\int \frac{1}{(x+1)\sqrt{x^2 - 1}}\) dx

• A. $\sqrt{\frac{x+1}{x-1}}+c$
• B. $\sqrt{\frac{x-1}{x+1}}+c$
• C. $\sqrt{\frac{1}{x+1}}+c$
• D. None of these

Hint:

Evaluate $\int\frac1(x+1)\sqrtx-1dx$

Answer 1

The Correct Option is B

Step 1: Concept
Use the substitution $x+1 = 1/t$.
Step 2: Analysis
Then $dx = -1/t^2 \, dt$. Substituting these into the integral leads to $-\int \frac{dt}{\sqrt{1-2t}}$.
Step 3: Evaluation
Integrating $-\int (1-2t)^{-1/2} \, dt$ gives $\sqrt{1-2t} + c$.
Step 4: Conclusion
Substituting $t = 1/(x+1)$ back, we get $\sqrt{1 - \frac{2}{x+1}} = \sqrt{\frac{x-1}{x+1}} + c$.
Final Answer: (b)