Deux problèmes sont présentés: le premier porte sur les transformations du plan, et le second sur l'étude d'une fonction et le calcul d'une aire.

The function is $f(x) = (\ln(1+x))^2$. This is a composite function of the form $u^2$, where $u(x) = \ln(1+x)$.
The derivative of $u(x)$ is $u'(x) = \frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x}$.

The derivative is defined for $1+x > 0$, which means $x > -1$. Thus, it is defined on $[0; +\infty[$.
The derivative of $f$ on $[0; +\infty[$ is $\boxed{\frac{2\ln(1+x)}{1+x}}$.

2. En déduire les variations de $f$ sur $[0; +\infty[$.

For $x \in [0; +\infty[$, we have $x \ge 0$, so $1+x \ge 1$. Therefore, $1+x > 0$.

For $x \in [0; +\infty[$, we have $1+x \ge 1$. Since the natural logarithm function $\ln(t)$ is non-negative for $t \ge 1$, it follows that $\ln(1+x) \ge \ln(1) = 0$.
Therefore, $2\ln(1+x) \ge 0$.

Since $2\ln(1+x) \ge 0$ and $1+x > 0$ for $x \in [0; +\infty[$, we have $f'(x) = \frac{2\ln(1+x)}{1+x} \ge 0$.
$f'(x) = 0$ if and only if $\ln(1+x) = 0$, which implies $1+x = 1$, so $x=0$.
As $f'(x) \ge 0$ on $[0; +\infty[$, the function $f(x)$ is increasing on this interval.

3. Calculer $f(0)$ et $f(1)$, puis dresser le tableau de variations de $f$ sur $[0; 1]$.

From part 2, $f'(x) \ge 0$ on $[0; +\infty[$, so $f'(x) \ge 0$ on $[0; 1]$. The function $f$ is increasing on $[0; 1]$.
The table of variations for $f$ on $[0; 1]$ is:

The values are $f(0) = \boxed{0}$ and $f(1) = \boxed{(\ln(2))^2}$.

4. Calculer $f'(0)$. En déduire l'équation de la tangente T à $\mathcal{C}$ au point d'abscisse 0.

Using the derivative $f'(x) = \frac{2\ln(1+x)}{1+x}$ from part 1:

So, $f'(0) = \boxed{0}$.

The equation of the tangent line at a point $(x_0, f(x_0))$ is given by $y - f(x_0) = f'(x_0)(x - x_0)$.
Here, $x_0 = 0$. We have $f(0) = 0$ (from part 3) and $f'(0) = 0$.

The equation of the tangent T is $\boxed{y=0}$.

5. Tracer dans $(o; \vec{i}, \vec{j})$ $\mathcal{C}$ en faisant apparaître la tangente T.

This step requires a graphical representation. Here are the key characteristics for drawing the graph:

• The curve $\mathcal{C}$ passes through the origin $(0,0)$.
• The tangent T at the origin is the x-axis ($y=0$). This means the curve is tangent to the x-axis at $(0,0)$.
• The function $f(x)$ is increasing on $[0; +\infty[$.
• The point $(1, f(1)) = (1, (\ln(2))^2)$ is on the curve. Approximately, $(\ln(2))^2 \approx (0.693)^2 \approx 0.48$.
• The graphical unit is $10 \text{ cm}$ for 1 unit. So, the point $(1, 0.48)$ would be $10 \text{ cm}$ along the x-axis and $4.8 \text{ cm}$ along the y-axis from the origin.
• As $x$ increases, $f(x)$ increases without bound.