$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$ equals
- A. $\frac{1}{2}\tan^{-1}\frac{\sqrt{2}x}{\sqrt{1-x^2}}$
- B. $\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{2}x}{\sqrt{1+x^2}}$
- C. $\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{2}x}{\sqrt{1-x^2}}$
- D. None of these
✅ Correct Answer: $\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{2}x}{\sqrt{1-x^2}}$
Explanation
Substitute $x=\sin\theta$, then simplify. The result simplifies to $(1/\sqrt{2})\tan^{-1}(\sqrt{2}x/\sqrt{1-x^2}) + c$.
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