If $y = \left(\log\left(x + \sqrt{x^2+a^2}\right)\right)^2$ and $x \neq \frac{1-a^2}{2}$, then $(x^2+a^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is equal to:

Given:

$y = \left[\log\left(x + \sqrt{x^2 + a^2}\right)\right]^2$

Step 1: Find the first derivative ($\frac{dy}{dx}$)

Using the chain rule:

$\frac{dy}{dx} = 2 \left[\log\left(x + \sqrt{x^2 + a^2}\right)\right] \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left(1 + \frac{2x}{2\sqrt{x^2 + a^2}}\right)$

Simplify the term inside the last parenthesis:

$1 + \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}$

Substitute this back into the derivative equation:

$\frac{dy}{dx} = 2 \left[\log\left(x + \sqrt{x^2 + a^2}\right)\right] \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{x + \sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2}}$

Canceling out the common term $x + \sqrt{x^2 + a^2}$:

$\frac{dy}{dx} = \frac{2 \log\left(x + \sqrt{x^2 + a^2}\right)}{\sqrt{x^2 + a^2}}$

Step 2: Rearrange and square both sides

Cross-multiply to remove the fraction:

$\sqrt{x^2 + a^2} \frac{dy}{dx} = 2 \log\left(x + \sqrt{x^2 + a^2}\right)$

Squaring both sides to eliminate the square root:

$(x^2 + a^2) \left(\frac{dy}{dx}\right)^2 = 4 \left[\log\left(x + \sqrt{x^2 + a^2}\right)\right]^2$

Since the right-hand side contains the original function $y$:

$(x^2 + a^2) \left(\frac{dy}{dx}\right)^2 = 4y$

Step 3: Differentiate implicitly with respect to $x$

Apply the product rule on the left side and the chain rule on the right side:

$\frac{d}{dx}(x^2 + a^2) \cdot \left(\frac{dy}{dx}\right)^2 + (x^2 + a^2) \cdot \frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^2\right] = 4\frac{dy}{dx}$

$2x \left(\frac{dy}{dx}\right)^2 + (x^2 + a^2) \cdot 2\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx^2}\right) = 4\frac{dy}{dx}$

Step 4: Simplify

Divide the entire equation by the common term $2\frac{dy}{dx}$:

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

Rearranging the terms matches the required expression perfectly:

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