Solution

$\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{\tan^{-1}x}{1 + x^2}$

I.F. $= e^{\tan^{-1}x}$

$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$

$y \cdot e^{\tan^{-1}x} = \tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} + c$

$y(0) = 1 \Rightarrow c = 2$

$y(1) = \frac{2}{e^4} + \frac{\pi}{4} - 1$