If $y=(x+\sqrt{x^2+1})^n$, then the value of $(x^2+1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}$ is:

n²y

$\frac{dy}{dx}=n(x+\sqrt{x^2+1})^{n-1}.[1+\frac{1}{2\sqrt{x^2+1}}.2x]$

$\frac{dy}{dx}=n(x+\sqrt{x^2+1})^{n-1}[\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}]=\frac{n(x+\sqrt{x^2+1})^{n}}{\sqrt{x^2+1}}$

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

Differentiating again w.r.t.x

$(x^2+1)2\frac{dy}{dx}.\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2.2x=x^22y.\frac{dy}{dx}$

Cancelling $2\frac{dy}{dx}$ on both side 

$(x^2+1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=n^2y$