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