The solution of differential equation $\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{1}{(1 + x^2)^2}$ is

$y(1 + x^2) = C + \tan^{-1} x$

The correct answer is Option (1) → $y(1 + x^2) = C + \tan^{-1} x$ ##

Given that, $\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{1}{(1 + x^2)^2} \Rightarrow \frac{dy}{dx} + P \cdot y = Q$

Here, $P = \frac{2x}{1 + x^2} \text{ and } Q = \frac{1}{(1 + x^2)^2}$

which is a linear differential equation.

$∴\text{I.F} = e^{\int \frac{2x}{1 + x^2} dx} \quad [∵\text{I.F} = e^{\int P dx}]$

Put $1 + x^2 = t \Rightarrow 2x \, dx = dt$

$∴\text{I.F} = e^{\int \frac{dt}{t}} = e^{\log t} = e^{\log(1+x^2)} = 1 + x^2$

The general solution is $y \cdot \text{I.F} = \int Q \cdot \text{I.F} \, dx + C$

$y \cdot (1 + x^2) = \int (1 + x^2) \cdot \frac{1}{(1 + x^2)^2} \, dx + C$

$\Rightarrow y(1 + x^2) = \int \frac{1}{1 + x^2} \, dx + C$

$\Rightarrow y(1 + x^2) = \tan^{-1} x + C$