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Question
\[\frac{dy}{dx} = \frac{1}{x^2 + 4x + 5}\]
Solution
We have,
\[\frac{dy}{dx} = \frac{1}{x^2 + 4x + 5}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x^2 + 4x + 4 + 1}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\left( x + 2 \right)^2 + \left( 1 \right)^2}\]
\[ \Rightarrow dy = \frac{1}{\left( x + 2 \right)^2 + \left( 1 \right)^2}dx\]
Integrating both sides, we get
\[\int y = \int\frac{1}{\left( x + 2 \right)^2 + \left( 1 \right)^2}dx\]
\[ \Rightarrow y = \tan^{- 1} \frac{x + 2}{1} + C\]
\[ \Rightarrow y = \tan^{- 1} \left( x + 2 \right) + C\]
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