Integrating factor of the differential equation \((1 + x^2)\dfrac{dy}{dx} + xy = 1\) is

The given differential equation is \((1 + x^2)\dfrac{dy}{dx} + xy = 1\)

Dividing the entire equation by \((1 + x^2)\), we get:

\(\dfrac{dy}{dx} + \dfrac{x}{1 + x^2}y = \dfrac{1}{1 + x^2}\)

This is a linear differential equation of the form \(\dfrac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = \dfrac{x}{1 + x^2}\) and \(Q(x) = \dfrac{1}{1 + x^2}\)

The integrating factor (IF) is given by \(e^{\int P(x) dx}\)

\(\int P(x) dx = \int \dfrac{x}{1 + x^2} dx\)

Let \(1 + x^2 = t\), then \(2x dx = dt \Rightarrow x dx = \dfrac{dt}{2}\)

\(\int \dfrac{x}{1 + x^2} dx = \dfrac{1}{2} \int \dfrac{dt}{t} = \dfrac{1}{2} \log t = \dfrac{1}{2} \log(1 + x^2) = \log(\sqrt{1 + x^2})\)

Therefore, the integrating factor is \(e^{\log(\sqrt{1 + x^2})} = \sqrt{1 + x^2}\)

Answer: \(\sqrt{1 + x^2}\)