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Question: Answered & Verified by Expert
$\int \frac{x}{\sqrt{1-2 x^4}} \mathrm{~d} x=$ (Where $C$ is a constant of integration)
MathematicsIndefinite IntegrationMHT CETMHT CET 2022 (08 Aug Shift 1)
Options:
  • A $\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{2} x^2\right)+C$
  • B $\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(2 \sqrt{2} x^2\right)+C$
  • C $\frac{1}{2} \sin ^{-1}(2 x)+C$
  • D $\frac{1}{\sqrt{2}} \sin ^{-1}(\sqrt{2} x)+C$
Solution:
1745 Upvotes Verified Answer
The correct answer is: $\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{2} x^2\right)+C$
$\int \frac{x}{\sqrt{1-2 x^4}} \mathrm{~d} x=\frac{1}{2 \sqrt{2}} \int \frac{2 \sqrt{2} x \mathrm{~d} x}{\sqrt{1-\left(\sqrt{2} x^2\right)^2}}=\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{2} x^2\right)$

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