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Question
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Solution
We have,
\[\left( x^2 + 1 \right)\frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x^2 + 1}\]
\[ \Rightarrow dy = \left( \frac{1}{x^2 + 1} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( \frac{1}{x^2 + 1} \right)dx\]
\[ \Rightarrow y = \tan^{- 1} x + C\]
\[\text{ So,} y = \tan^{- 1} x + \text{ C is defined for all }x \in R . \]
\[\text{ Hence, }y = \tan^{- 1} x + \text{ C, where }x \in R, \text{ is the solution to the given differential equation }.\]
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