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$\int x\text{Tan}^{-1}\sqrt{\frac{1+x^2}{1-x^2}}dx=$

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For integrals involving terms like $\sqrt{\frac{1 \pm x^2}{1 \mp x^2}}$, the substitution $x^2 = \cos\theta$ is often very effective, as it allows for simplification using half-angle trigonometric identities.

Updated On: Mar 30, 2026

$\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)+\frac{1}{4}\sqrt{1-x^4}+c$
$\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^4}+c$
$\frac{x^2}{4}(\pi+\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^4}+c$
$\frac{x^2}{4}(\pi+\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^2}+c$

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The Correct Option is A

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$\int x\text{Tan}^{-1}\sqrt{\frac{1+x^2}{1-x^2}}dx=$

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The Correct Option is A

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