Question

\(\int\frac{dx}{(x+1)\sqrt{x^2-1}}\) is equal to:
[\(c\) is an arbitrary constant]

• A. \(\sqrt{\frac{x+1}{x-1}}+c\)
• B. \(\sqrt{\frac{x}{x-1}}+c\)
• C. \(\sqrt{\frac{x-1}{x+1}}+c\)
• D. \(\sqrt{\frac{x}{x+1}}+c\)

Hint:

Write \(\sqrt{x^2-1}=\sqrt{(x-1)(x+1)}\) and substitute \(t=\frac{x-1}{x+1}\).

Answer 1

The Correct Option is C

Step 1: Approach:
Try the substitution $x=\sec\theta$, then simplify with half-angle identities.

Step 2: Substitute:
Let $x=\sec\theta$. Then $dx=\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2-1}=\tan\theta$. The integral becomes $\int\frac{\sec\theta\tan\theta}{(\sec\theta+1)\tan\theta}d\theta=\int\frac{\sec\theta}{\sec\theta+1}d\theta$.

Step 3: Simplify:
$\frac{\sec\theta}{\sec\theta+1}=\frac{1}{1+\cos\theta}=\frac{1}{2\cos^2(\theta/2)}$. So the integral is $\int\frac12\sec^2\frac{\theta}{2}d\theta=\tan\frac{\theta}{2}+c$.

Step 4: Return to x:
$\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$. With $\cos\theta=\frac1x$ this is $\sqrt{\frac{x-1}{x+1}}$.

Step 5: Choose:
The result $\sqrt{\frac{x-1}{x+1}}+c$ is option 3.

Final Answer:
\[ \boxed{\sqrt{\frac{x-1}{x+1}}+c} \]