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Question
\[\frac{dy}{dx} = y^2 + 2y + 2\]
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Solution
We have,
\[\frac{dy}{dx} = y^2 + 2y + 2\]
\[ \Rightarrow \frac{dy}{dx} = y^2 + 2y + 1 + 1\]
\[ \Rightarrow \frac{dy}{dx} = \left( y + 1 \right)^2 + 1^2 \]
\[ \Rightarrow \frac{1}{\left( y + 1 \right)^2 + \left( 1 \right)^2}dy = dx\]
Integrating both sides, we get
\[\int\frac{1}{\left( y + 1 \right)^2 + \left( 1 \right)^2}dy = \int dx\]
\[ \Rightarrow \tan^{- 1} \left( \frac{y + 1}{1} \right) + C = x\]
\[ \Rightarrow x = \tan^{- 1} \left( y + 1 \right) + C\]
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