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प्रश्न
उत्तर
\[\int\frac{dx}{x\sqrt{x^4 - 1}}\]
` = ∫ {x dx}/{x^2 \sqrt{( x^2 )^2 - 1} `
` Let x^2 = t `
\[ \Rightarrow 2x = \frac{dt}{dx}\]
\[ \Rightarrow \text{x dx} = \frac{dt}{2}\]
Now, ` = ∫ {x dx}/{x^2 \sqrt{( x^2 )^2 - 1} `
\[ = \frac{1}{2}\int\frac{dt}{t\sqrt{t^2 - 1}}\]
\[ = \frac{1}{2} \sec^{- 1} \left( t \right) + C\]
\[ = \frac{1}{2} \sec^{- 1} \left( x^2 \right) + C\]
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