If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$. 
If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.

My try: $$F_Y(x) = P(X^2+1 < x) = \mathrm P(-\sqrt{x-1} < X < \sqrt{x-1}) = \mathrm P(X < \sqrt{x-1}) -\mathrm  P(X < -\sqrt{x-1})$$
Now since $$F_X(x) = \begin{cases}\ 1-e^{-\lambda x}, & x \ge 0 \\ 0 , & x< 0\end{cases}$$
we have $\mathrm  P(X < -\sqrt{x-1}) \equiv 0$. Hence
$$F_Y(x) = \mathrm P(X < \sqrt{x-1}) = F_X(\sqrt{x-1}) = \begin{cases} 1-e^{-\lambda \sqrt{x-1}}, & \sqrt{x-1} \ge 0 \\ 0 , & \sqrt{x-1}< 0\end{cases}$$
or
$$F_Y(x) = \begin{cases} 1- e^{-\lambda \sqrt{x-1}}, & x > 1 \\ 0 , & x \le 1\end{cases}$$

*

*I'm not sure putting the last intervals for $x$ though. Is there a mistake in this solution?


*Now, I'm going to differentiate $F_Y(x)$ to find the probability density function. Is there any shortcut/other approach to this?

\begin{align}
F_Y(x) & = \Pr(X^2+1\le x) \\[8pt]
& = \begin{cases} \Pr(X\le \sqrt{x-1} ) & \text{if } x\ge1, \\
0 & \text{if } x<1
\end{cases} \\[8pt]
& = \begin{cases}
1 - e^{-\lambda\sqrt{x-1}} & \text{if } x\ge 1 \\
0 & \text{if } x<1 \quad \text{(not “if $\sqrt{x-1} < 0$”)}
\end{cases}
\end{align}
