1. Let $y=\left(x+\sqrt{1+x^2}\right)^n.$ We need to find $\left(1+x^2\right)\frac{d^2y}{dx^2}+x\frac{dy}{dx}.$

2. Take logarithm: $\ln y=n\ln\left(x+\sqrt{1+x^2}\right).$ Differentiate both sides: $\frac{1}{y}\frac{dy}{dx}=n\cdot \frac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right).$

Now, \frac{d}{dx}\left(x+\sqrt{1+x^2}\right)=1+\frac{x}{\sqrt{1+x^2}}=rac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}. Hence, $\frac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right)=\frac{1}{x+\sqrt{1+x^2}}\cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}.$ So, $\frac{y'}{y}=\frac{n}{\sqrt{1+x^2}}$ which gives $y'=\frac{ny}{\sqrt{1+x^2}}.$

3. Differentiate again: $y''=\frac{d}{dx}\left(\frac{ny}{\sqrt{1+x^2}}\right).$ Using product rule, $y''=n\left[\frac{y'}{\sqrt{1+x^2}}+y\frac{d}{dx}(1+x^2)^{-1/2}\right].$ Now, $\frac{d}{dx}(1+x^2)^{-1/2}=-\frac{x}{(1+x^2)^{3/2}},$ and $y'=\frac{ny}{\sqrt{1+x^2}}.$ Therefore, $y''=n\left[\frac{ny}{1+x^2}-\frac{xy}{(1+x^2)^{3/2}}\right].$ So, $y''=\frac{n^2y}{1+x^2}-\frac{nxy}{(1+x^2)^{3/2}}.$

4. Now compute the required expression: $\left(1+x^2\right)y''+xy'.$ Substitute $y''$ and $y'$: $\left(1+x^2\right)\left(\frac{n^2y}{1+x^2}-\frac{nxy}{(1+x^2)^{3/2}}\right)+x\left(\frac{ny}{\sqrt{1+x^2}}\right).$ Simplify: $=n^2y-\frac{nxy}{\sqrt{1+x^2}}+\frac{nxy}{\sqrt{1+x^2}}=n^2y.$

5. Hence, $\boxed{\left(1+x^2\right)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=n^2y.}$

6. Option check:

• A: $n^2y$ ✅
• B: $-n^2y$ ❌
• C: $-y$ ❌
• D: $2x^2y$ ❌

Therefore, the correct option is A.