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प्रश्न

\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
योग

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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अध्याय 19: Indefinite Integrals - Exercise 19.18 [पृष्ठ ९९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.18 | Q 6 | पृष्ठ ९९

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