Indefinite Integration
Integration by Parts
Grade None
Question:
<p>[JEE Main 2022] \(\displaystyle\int\sin^{-1}\!\frac{2x}{1+x^2}\,dx\) equals (for \(|x|\le 1\))</p>
<li>\(2x\tan^{-1}x-\ln(1+x^2)+C\)</li>
<li>\(2\!\left[x\tan^{-1}x-\ln(1+x^2)\right]+C\)</li>
<li>\(2x\tan^{-1}x+\ln(1+x^2)+C\)</li>
<li>\(x\tan^{-1}x-\ln(1+x^2)+C\)</li>
Step-by-Step Solution
Key Concept: Since sin⁻^1(2x/(1+x^2)) = 2tan⁻^1x for |x|\leq1, the integral becomes 2\inttan⁻^1x dx. Integrate by parts.
<p><strong>Simplify:</strong> For $|x|\le 1$, $\sin^{-1}\!\dfrac{2x}{1+x^2}=2\tan^{-1}x$ (double angle for sine).</p>
<p>$$I = 2\int\tan^{-1}x\,dx$$</p>
<p><strong>By parts:</strong> $u=\tan^{-1}x,\;dv=dx\Rightarrow v=x$.</p>
<p>$$= 2\!\left[x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx\right]+C = 2\!\left[x\tan^{-1}x-\frac12\ln(1+x^2)\right]+C$$</p>
<p>$$= 2x\tan^{-1}x-\ln(1+x^2)+C$$</p>
<p>Answer: <strong>(B)</strong></p>
Correct Answer: B