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