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