Could you please help to solve the following problem:
The random variable $X$ has a uniform distribution on the segment $[0,2]$. Find the density of the distribution of a random variable $Y = (X + 1)^2$.
My attempt (but it is obviously wrong):
$$ f(x) = \begin{cases} (x+1)^2, & x \notin [0,2], \\ 0, & x \notin [0,2]. \end{cases} $$
$$ \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x = \int_{0}^{2} (x+1)^2 \, \mathrm{d}x = \left[ (x+1)^2 \right]_{0}^{2} = (x+1)^2 (2 - 0) = 1. $$
\begin{align*} &\implies \quad 2(x+1)^2 = 1 \\[0.5em] &\implies \quad (x+1)^2 = \frac{1}{2} \\[0.5em] &\implies \quad x^2 + 2x + \frac{1}{2} = 0 \\[0.5em] &\implies \quad x_1 = \frac{-2 + 2^{1/2}}{2}, \quad x_2 = \frac{-2 - 2^{1/2}}{2} \end{align*}
$$ P (Y \leq u) = P((X+1)^2 \leq u) = P(X \leq \sqrt{u}-1)$$ Thus, using $F$ for the cummulative distribution function we have $$F_Y (u) = F_x (\sqrt{u}-1)$$ The density $f$ is a derivative of $F$: $$f_Y(u) = \frac{d}{du} F_X(\sqrt{u}-1) = F_X^\prime (\sqrt{u}-1) \frac{1}{2\sqrt{u}}$$ $$ = \frac{f_x\left(\sqrt{u}-1 \right)}{2\sqrt{u}} = \left\{ \begin{array}{ccc} \frac{1}{4\sqrt{u}} & \;\;if\;\; & 1 \leq u \leq 9 \\ 0 & \;\; elsewhere \end{array} \right. $$