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प्रश्न
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उत्तर
` ∫ {x dx }/ {x^4 + 2 x^2 +3} `
\[\text{ let } x^2 = t\]
\[ \Rightarrow \text{ 2x dx }= dt\]
\[ \Rightarrow \text{ x dx }= \frac{dt}{2}\]
Now, ` ∫ {x dx }/ {x^4 + 2 x^2 +3} `
\[ = \frac{1}{2}\int\frac{dt}{t^2 + 2t + 3}\]
\[ = \frac{1}{2}\int\frac{dt}{t^2 + 2t + 1 + 2}\]
\[ = \frac{1}{2}\int\frac{dt}{\left( t + 1 \right)^2 + \left( \sqrt{2} \right)^2} \]
\[ = \frac{1}{2} \times \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{t + 1}{\sqrt{2}} \right) + C \left[ \because \int\frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{- 1} \left( \frac{x}{a} \right) + C \right]\]
\[ = \frac{1}{2\sqrt{2}} \tan^{- 1} \left( \frac{x^2 + 1}{\sqrt{2}} \right) + C\]
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