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प्रश्न
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उत्तर
We have,
\[\frac{dy}{dx} = \left( x + y + 1 \right)^2 \]
\[\text{ Putting }x + y + 1 = v\]
\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
\[ \therefore \frac{dv}{dx} - 1 = v^2 \]
\[ \Rightarrow \frac{dv}{dx} = v^2 + 1\]
\[ \Rightarrow \frac{1}{v^2 + 1}dv = dx\]
Integrating both sides, we get
\[\int\frac{1}{v^2 + 1}dv = \int dx\]
\[ \Rightarrow \tan^{- 1} v = x + C\]
\[ \Rightarrow \tan^{- 1} \left( x + y + 1 \right) = x + C\]
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