If $X$ is distributed $\mathcal{N}\sim(\mu,1)$ then what is the PDF of $X^2$? If $X$ is distributed $\mathcal{N}\sim(\mu,1)$ then what is the distribution of $X^2$?  Here is my try:
$$\begin{align*}
\mathbb{P}(X^2<x) 
&= \mathbb{P}(-\sqrt{x}<X<\sqrt{x}) \\[6pt]
&=  \int_{-\sqrt{x}}^{\sqrt{x}} f(x) \, dx. \\[6pt]
\end{align*}$$
Then:
$$\begin{align*}
f_{X^{2}}(x)
&= \frac{d}{dx}\int_{\sqrt{-x}}^{\sqrt{x}} f(x) \,dx \\[6pt]
&= f_X(\sqrt{x})\frac{1}{2\sqrt{x}}+f_X(-\sqrt{x})\frac{1}{2\sqrt{x}} \\[6pt]
&= \frac{1}{\sqrt{2\pi x}} e^{\frac{-(x-\mu)^2}{2}}. \\[6pt]
\end{align*}$$
How should I continue?
 A: Your algebra is wrong in the transition from your penultimate step to your last step (see correct step below).  Part of the problem here is that you are using $x$ both as the variable-of-integration and as the argument for your new density function, which confuses your mathematical statements.  To simplify the notation, let us instead take $R = X^2$ as the new random variable of interest, and specify its arguments by $r$.  You then get:
$$\mathbb{P}(R \leqslant r) = \mathbb{P}(- \sqrt{r} \leqslant X \leqslant \sqrt{r})
= \int \limits_{-\sqrt{r}}^{\sqrt{r}} f_X(x) \ dx.$$
Using Leibniz integral rule on your probability integral you should then get (omitting some of the steps, which I will leave for you):
$$\begin{align}
f_R(r) 
&= \frac{d}{dr} \int \limits_{-\sqrt{r}}^{\sqrt{r}} f_X(x) \ dx \\[6pt]
&= \frac{1}{2 \sqrt{r}} \cdot f_X(\sqrt{r}) + \frac{1}{2 \sqrt{r}} f_X(-\sqrt{r}) \\[6pt]
&= \frac{1}{2 \sqrt{r}} \Bigg[ \text{N}(\sqrt{r}|\mu, 1) + \text{N}(-\sqrt{r}|\mu, 1) \Bigg] \\[6pt]
&= \frac{1}{2 \sqrt{r}} \cdot \frac{1}{\sqrt{2 \pi}} \Bigg[ \exp \Big(-\frac{1}{2} ( \sqrt{r} - \mu)^2 \Big) + \exp \Big(-\frac{1}{2} ( -\sqrt{r} - \mu)^2 \Big) \Bigg] \\[6pt]
&\ \ \ \vdots \\[6pt]
&= \frac{1}{\sqrt{2 \pi r}} \cdot \exp \Big( - \frac{r + \mu^2}{2} \Big) \cdot \cosh(\mu \sqrt{r}) \\[12pt]
&= \text{ChiSq}(r | \text{df} = 1, \text{ncp} = \mu^2). \\[12pt]
\end{align}$$
This demonstrates that the random variable $R=X^2$ has a non-central chi-squared distribution with one-degree-of-freedom and non-centrality parameter $\mu^2$.  This is consistent with the general theory for how this kind of distribution arises in general.
