To solve the differential equation \((x^2+1)\frac{dy}{dx} + 2xy = \sqrt{x^2+4}\), we can use the method of integrating factors for a linear first-order differential equation.

The given differential equation can be written in the standard form:

\(\frac{dy}{dx} + P(x)y = Q(x)\)

where \(P(x) = \frac{2x}{x^2+1}\) and \(Q(x) = \frac{\sqrt{x^2+4}}{x^2+1}\).

The integrating factor \(\mu(x)\) is given by:

\(\mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2x}{x^2+1} \, dx}\)

Calculating the integral:

\(\int \frac{2x}{x^2+1} \, dx = \ln |x^2+1|\)

Therefore, the integrating factor is:

\(\mu(x) = e^{\ln |x^2+1|} = |x^2+1|\)

Since \(x^2 + 1\) is always positive, we have:

\(\mu(x) = x^2+1\)

Multiply the entire differential equation by the integrating factor:

\((x^2+1)\frac{dy}{dx} + 2x(x^2+1)y = \sqrt{x^2+4}\)

The left side of the equation becomes the derivative of \(((x^2+1)y)\):

\(\frac{d}{dx}((x^2+1)y) = \sqrt{x^2+4}\)

Integrate both sides with respect to \(x\):

\int \sqrt{x^2+4} \, dx, perform a trigonometric substitution (e.g., \(x = 2\tan(\theta)\)):

This simplifies to:

y = (x^2+1)^{-1}(\frac{1}{2}x\sqrt{x^2+4}+2\ln |x+\sqrt{x^2+4}|) + C

Thus, the correct answer is:

\(y=(x^2+1)^{-1}(\frac{1}{2}x\sqrt{x^2+4}+2log|(x+\sqrt{x^2+4})|) + c; \text{ where c is a constant}\)