Do not ever send me answers/solutions in french again. If not I will never use this scan solve again

The function is $y = \ln(x^2 \sqrt{x^2+1})$.
Using the logarithm property $\ln(ab) = \ln a + \ln b$:

$$y = \ln(x^2) + \ln(\sqrt{x^2+1})$$

Rewrite the square root as a power: $\sqrt{x^2+1} = (x^2+1)^{1/2}$.

$$y = \ln(x^2) + \ln((x^2+1)^{1/2})$$

Using the logarithm property $\ln(a^b) = b \ln a$:

$$y = 2 \ln x + \frac{1}{2} \ln(x^2+1)$$

Recall the differentiation rule for natural logarithms: $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$.

Differentiate the first term, $2 \ln x$:

$$\frac{d}{dx}(2 \ln x) = 2 \cdot \frac{1}{x} = \frac{2}{x}$$

Differentiate the second term, $\frac{1}{2} \ln(x^2+1)$. Here, $u = x^2+1$, so $\frac{du}{dx} = 2x$.

$$\frac{d}{dx}\left(\frac{1}{2} \ln(x^2+1)\right) = \frac{1}{2} \cdot \frac{1}{x^2+1} \cdot (2x)$$ $$= \frac{2x}{2(x^2+1)} = \frac{x}{x^2+1}$$

Find a common denominator, which is $x(x^2+1)$.

$$\frac{dy}{dx} = \frac{2(x^2+1)}{x(x^2+1)} + \frac{x \cdot x}{x(x^2+1)}$$ $$\frac{dy}{dx} = \frac{2x^2+2 + x^2}{x(x^2+1)}$$ $$\frac{dy}{dx} = \frac{3x^2+2}{x(x^2+1)}$$

The final answer is $\frac{3x^2+2}{x(x^2+1)}$.