Answer
$\color{blue}{\displaystyle \left(\frac{1-e^t}{t}\right)^2}$
Work Step by Step
$Y$ has pdf:
$f_Y(y) = \begin{cases} y,& 0\le y\le 1\\ 2-y,& 1\le y\le 2 \\ 0, & \text{elsewhere} \end{cases}$
$\begin{align*}
M_Y(t) &= E(e^{Yt}) \\
&= \int_{-\infty}^\infty e^{yt}f_Y(y)\ dy \\
&= \int_0^1 ye^{yt}\ dy + \int_1^2 (2-y)e^{yt}\ dy \qquad \text{[ since}\ f_Y(y)=0, y\not\in [1,2]\ ] \\
&= \underbrace{\int_0^1 ye^{yt}\ dy}_{\text{IBP}} + \int_1^2 2e^{yt}\ dy - \underbrace{\int_1^2 ye^{yt}\ dy}_{\text{IBP}} \\
& \\
& \qquad\text{Use integration by parts (IBP):} \\
& \qquad \begin{array}{l|l} u = y & dv = e^{yt}\ dy \\ \hline du = dy & \displaystyle v = \frac{e^{yt}}{t} \end{array} \\
& \qquad \int u\ dv = uv-\int v\ du \\
& \qquad \int ye^{yt}\ dy = y\frac{e^{yt}}{t} - \int \frac{e^{yt}}{t}\ dy \\
& \\
&= \Biggl(\left( y\frac{e^{yt}}{t}\ \right\vert_0^1 - \int_0^1 \frac{e^{yt}}{t}\ dy \Biggr) + \left( 2\frac{e^{yt}}{t}\ \right\vert_1^2 + \Biggl(\left( y\frac{e^{yt}}{t}\ \right\vert_1^2 - \int_1^2 \frac{e^{yt}}{t}\ dy\Biggr) \\
&= \Biggl(\left(1\frac{e^t}{t}-0\frac{e^{0t}}{t}\right) - \left(\frac{e^{yt}}{t^2}\ \right\vert_0^1 \Biggr) + 2\left(\frac{e^{2t}}{t} - \frac{e^{t}}{t} \right) \\
& \qquad\qquad\qquad\qquad\qquad - \Biggl(\left(2\frac{e^{2t}}{t} - 1\frac{e^{t}}{t}\right) - \left(\frac{e^{yt}}{t^2}\ \right\vert_1^2 \Biggr)\\
&= \Biggl( \frac{e^t}{t} - \left(\frac{e^{t}}{t^2} - \frac{1}{t^2}\right) \Biggr) + \left(\frac{2(e^{2t}-e^t)}{t}\right) \\
& \qquad\qquad\qquad\qquad\qquad - \Biggl(\left(\frac{e^{2t}}{t} - \frac{e^{t}}{t}\right) - \left(\frac{e^{2t}}{t^2} + \frac{e^t}{t^2} \right) \Biggr)\\
&= \frac{te^t -e^t+1}{t^2} + \frac{2te^{2t}-2te^t}{t^2} - \frac{2te^{2t} - te^t -e^{2t} + e^t}{t^2} \\
&= \frac{te^t -e^t+1}{t^2} + \frac{2te^{2t}-2te^t}{t^2} + \frac{-2te^{2t} + te^t +e^{2t} - e^t}{t^2} \\
&= \frac{e^{2t} -2e^t +1 }{t^2} \\
&= \frac{(1-e^t)^2}{t^2} \\
\color{blue}{M_Y(t)}\ &\color{blue}{= \left(\frac{1-e^t}{t}\right)^2}
\end{align*}$
