Question:
Evaluate $\int \frac{1+x^2}{\sqrt{1-x^2}} dx$:
Evaluate $\int \frac{1+x^2}{\sqrt{1-x^2}} dx$:
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Use substitution for $\sqrt{1-x^2}$ type integrals.
Updated On: Jun 10, 2026
- $\frac{3}{2}\sin^{-1}x - \frac{x}{2}\sqrt{1-x^2}+c$
- $\frac{3}{2}\sin^{-1}x + \frac{x}{2}\sqrt{1-x^2}+c$
- $\frac{1}{2}\sin^{-1}x - \frac{x}{2}\sqrt{1-x^2}+c$
- $\frac{1}{2}\sin^{-1}x + \frac{x}{2}\sqrt{1-x^2}+c$
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The Correct Option is A
Solution and Explanation
Let $x=\sin\theta$
\[
I=\int (1+\sin^2\theta)d\theta
\]
\[
I=\frac{3}{2}\theta - \frac{1}{2}\sin\theta\cos\theta
\]
\[
I=\frac{3}{2}\sin^{-1}x - \frac{x}{2}\sqrt{1-x^2}
\]
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