Question

\[\frac{dy}{dx} = \tan^{- 1} x\]

Answer 1

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

\[ \Rightarrow y = \tan^{- 1} x\int1 dx - \int\left[ \frac{d}{dx}\left( \tan^{- 1} x \right)\int1 dx \right]dx\]
\[ \Rightarrow y = x \tan^{- 1} x - \int\frac{x}{1 + x^2}dx\]
\[ \Rightarrow y = x \tan^{- 1} x - \frac{1}{2}\int\frac{2x}{1 + x^2}dx\]
\[ \Rightarrow y = x \tan^{- 1} x - \frac{1}{2}\log\left| 1 + x^2 \right| + C\]
\[\text{ So, } y = x \tan^{- 1} x - \frac{1}{2}\log\left| 1 + x^2 \right| +\text{C is defined for all }x \in R.\]
\[\text{ Hence, } y = x \tan^{- 1} x - \frac{1}{2}\log\left| 1 + x^2 \right| +\text{C is the solution to the given differential equation.}\]