Question:
The integral $\int \sin^{-1}\left(\frac{2x}{1 + x^2}\right)\,dx$ for $|x| < 1$ is equal to
The integral $\int \sin^{-1}\left(\frac{2x}{1 + x^2}\right)\,dx$ for $|x| < 1$ is equal to
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Whenever you integrate a standalone inverse function, remember that it implicitly requires Integration by Parts where the second hidden function is simply $1$. The first term of the result will always feature $x$ multiplied directly by that inverse function, which immediately narrows down your choices in this problem to option (D)!
Updated On: Jun 18, 2026
- $2\tan^{-1}x - \log|1 + x^2| + c$
- $x\tan^{-1}x + \log|1 + x^2| + c$
- $\tan^{-1}x + \log|1 + x^2| + c$
- $2x\tan^{-1}x - \log|1 + x^2| + c$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the indefinite integral of an inverse trigonometric expression containing a rational function argument within the domain constraint $|x| < 1$.
Step 2: Key Formula or Approach:
1. Apply the standard trigonometric identity: for $|x| < 1$, $\sin^{-1}\left(\frac{2x}{1 + x^2}\right) = 2\tan^{-1}x$. 2. Integrate the simplified expression using Integration by Parts (ILATE rule): $$\int u \cdot v \, dx = u \int v \, dx - \int \left( u' \int v \, dx \right) dx$$
Step 3: Detailed Explanation:
Let's rewrite the integral using the identity $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$: $$I = \int 2\tan^{-1}x \, dx = 2 \int (\tan^{-1}x \cdot 1) \, dx$$ Apply integration by parts with $u = \tan^{-1}x$ and $v = 1$: Derivative of $u$: $u' = \frac{1}{1 + x^2}$ Integral of $v$: $\int 1 \, dx = x$ Expanding the formula: $$I = 2 \left[ \left(\tan^{-1}x\right) \cdot x - \int \frac{1}{1 + x^2} \cdot x \, dx \right]$$ $$I = 2x\tan^{-1}x - 2 \int \frac{x}{1 + x^2} \, dx$$ To solve the remaining integral, use a basic substitution where $t = 1 + x^2$, which means $dt = 2x \, dx$: $$2 \int \frac{x}{1 + x^2} \, dx = \int \frac{2x}{1 + x^2} \, dx = \int \frac{1}{t} \, dt = \log|t| = \log|1 + x^2|$$ Combine everything together and add the constant of integration $c$: $$I = 2x\tan^{-1}x - \log|1 + x^2| + c$$
Step 4: Final Answer:
The expression matches exactly with option (D).
The question asks for the indefinite integral of an inverse trigonometric expression containing a rational function argument within the domain constraint $|x| < 1$.
Step 2: Key Formula or Approach:
1. Apply the standard trigonometric identity: for $|x| < 1$, $\sin^{-1}\left(\frac{2x}{1 + x^2}\right) = 2\tan^{-1}x$. 2. Integrate the simplified expression using Integration by Parts (ILATE rule): $$\int u \cdot v \, dx = u \int v \, dx - \int \left( u' \int v \, dx \right) dx$$
Step 3: Detailed Explanation:
Let's rewrite the integral using the identity $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$: $$I = \int 2\tan^{-1}x \, dx = 2 \int (\tan^{-1}x \cdot 1) \, dx$$ Apply integration by parts with $u = \tan^{-1}x$ and $v = 1$: Derivative of $u$: $u' = \frac{1}{1 + x^2}$ Integral of $v$: $\int 1 \, dx = x$ Expanding the formula: $$I = 2 \left[ \left(\tan^{-1}x\right) \cdot x - \int \frac{1}{1 + x^2} \cdot x \, dx \right]$$ $$I = 2x\tan^{-1}x - 2 \int \frac{x}{1 + x^2} \, dx$$ To solve the remaining integral, use a basic substitution where $t = 1 + x^2$, which means $dt = 2x \, dx$: $$2 \int \frac{x}{1 + x^2} \, dx = \int \frac{2x}{1 + x^2} \, dx = \int \frac{1}{t} \, dt = \log|t| = \log|1 + x^2|$$ Combine everything together and add the constant of integration $c$: $$I = 2x\tan^{-1}x - \log|1 + x^2| + c$$
Step 4: Final Answer:
The expression matches exactly with option (D).
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