Density of $ Y = X + \frac{1}{X}$ when $X\sim U(a,b)$ Let $ X $ be a continuous random variable uniformly distributed in $ \left[a, b\right] $, that is $X \sim U(a, b)$. Suppose $ 0 < a < b $.
We wish to find the density function of $$ Y = X + \frac{1}{X} $$ Therefore, we want $ f_Y(y) $.
It is obvious that for $$ y\in (-2, 2) => f_Y(y) = P(Y = y) = P\left(X + \frac{1}{X} = y\right) = 0 $$ because $X$ takes real values, and therefore it is known that the function $ f(x) = x + \frac{1}{x} $ ranges in $(-\infty, -2]\cup[2, \infty)$.
I tried using the definition for $y\in(-\infty, -2]\cup[2, \infty)$: $$ P(Y = y) = P\left(X + \frac{1}{X} = y\right) = P(X^2 - yX + 1 = 0) = \cdots = \\P\left(\left(X - \frac{y + \sqrt{y^2 - 4}}{2} \right)\left(X - \frac{y - \sqrt{y^2 - 4}}{2} \right) = 0\right) = \\
P\left(X = \frac{y + \sqrt{y^2 - 4}}{2} \right) + P\left(X = \frac{y - \sqrt{y^2 - 4}}{2} \right) $$ which seems awfully dull. I am not even sure if it works.
I tried considering the random variable $Z = \frac{1}{X}$.  It can easily be derived that $ f_Z(z) = \frac{z^{-2}}{b -a} $ for $b^{-1} < z < a^{-1} $. Therefore we can say $ Y = X + Z $.  From here we can proceed:
$$ P(Y = y) = P(X + Z = y) = \int_{0}^{y} P(X = y - k, Z = k)dk $$
But firstly the random variables $X$ and $Z$ are dependant, and secondly, it seems wrong. Any ideas?

A possible path
Let $F_X(x)$ and $f_X(x)$ be the c.d.f and p.d.f. of $X\sim\mathcal U(a,b)$. Let also
$$g(x) = \frac1x + x,$$
$$\alpha(y) = \frac{y-\sqrt{y^2-4}}{2},$$
$$\beta(y) = \frac{y+\sqrt{y^2-4}}{2}.$$
Case 1 ($a\geq 1$)
$g(x)$ in monotonically increasing in $[a,b]$. Thus
\begin{eqnarray}
P(Y<y) &=& \begin{cases}
P(X<\beta(y)) & (g(a) \leq y \leq g(b))\\
0 & \mbox{(otherwise)}
\end{cases}\\
&=&\begin{cases}
F_X(\beta(y)) & (g(a) \leq y \leq g(b))\\
0 & \mbox{(otherwise)}
\end{cases}
\end{eqnarray}
Consequently, by differentiation you derive the p.d.f. of $Y=\frac1X+X$ as
\begin{eqnarray}
f_Y(y) &=& \begin{cases}
f_X(\beta(y)) \cdot \frac{d\beta}{d y} & (g(a) \leq y \leq g(b))\\
0 &\mbox{(otherwise)}
\end{cases}\\
&=&\begin{cases}
\frac1{b-a}\left(
\frac12+\frac{y}{2\sqrt{y^2-4}}\right) & (g(a) \leq y \leq g(b))\\
0 &\mbox{(otherwise)}
\end{cases}
\end{eqnarray}
Case 2 ($b\leq 1$)
You proceed in a very similar manner, by noting that now $g(x)$ in monotonically decreasing in $[a,b]$. Thus
\begin{eqnarray}
P(Y<y) = \begin{cases}
P(X<\alpha(y)) & (g(b) \leq y \leq g(a))\\
0 & \mbox{(otherwise)}
\end{cases}
\end{eqnarray}
Case 3 ($0<a < 1 <b$ and $ab>1$)
In this case you have $g(b) > g(a)>2$, so
$$
P(Y<y) =
\begin{cases}
P(\alpha(y) <x< \beta(y)) & (2<y< g(a))\\
P(x<\beta(y)) & (g(a)<y<g(b))\\
0 & (\mbox{otherwise})
\end{cases}
$$
Case 4 ($0<a < 1 <b$ and $0<ab<1$)
In this case you have $2<g(b) < g(a)$, so
$$
P(Y<y) =
\begin{cases}
P(\alpha(y) <x< \beta(y)) & (2<y< g(b))\\
P(x<\alpha(y)) & (g(b)<y<g(a))\\
0 & (\mbox{otherwise})
\end{cases}
$$
