Use Laplace transform to solve initial value prob. The problem is: $y" + 9y = e^t$, with the initial conditions $y(0) = 0, y'(0) = 0$.

I'm stuck at the inverse Laplace transform part.  Do I have to use partial fraction expansion or can I just split the equation $\frac{1}{(s-1)(s^2+9)}$?
You do not have to use partial fractions. I have no idea though about your meaning of split the equation. I think most people would infer splitting the equation as partial fractions decomposition. If you don't want to use partial fractions, we can use the inverse Laplace transform definition. That is,
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds = \sum_j\operatorname{Res}\{F(s);s_j\}\tag{1}
$$
where $F(s) = \frac{1}{(s-1)(s^2 + 9)}$. The poles of $F(s)$ occur when $s = 1, \pm 3i$; thus, we only have simple poles. Using equation $(1)$, we have
\begin{align}
\frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s-1)(s^2+9)}ds &amp; = \lim_{s\to 1}(s-1)\frac{e^{st}}{(s-1)(s^2+9)}+\lim_{s\to 3i}(s-3i)\frac{e^{st}}{(s-1)(s^2+9)}\\
&amp;+\lim_{s\to -3i}(s+3i)\frac{e^{st}}{(s-1)(s^2+9)}\\
&amp;= \lim_{s\to 1}\frac{e^{st}}{s^2+9}+\lim_{s\to 3i}\frac{e^{st}}{(s-1)(s+3i)}+\lim_{s\to -3i}\frac{e^{st}}{(s-1)(s-3i)}\\
&amp;\tag{2}
\end{align}
After taking the limit in equation $(2)$ and simplifying, you should get 
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{30}\bigl[3e^t - 3\cos(3t)-\sin(3t)\bigr]
$$
You do not have to use partial fractions. I have no idea though about your meaning of split the equation. I think most people would infer splitting the equation as partial fractions decomposition. If you don't want to use partial fractions, we can use the inverse Laplace transform definition. That is,
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds = \sum_j\operatorname{Res}\{F(s);s_j\}\tag{1}
$$
where $F(s) = \frac{1}{(s-1)(s^2 + 9)}$. The poles of $F(s)$ occur when $s = 1, \pm 3i$; thus, we only have simple poles. Using equation $(1)$, we have
\begin{align}
\frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s-1)(s^2+9)}ds &amp; = \lim_{s\to 1}(s-1)\frac{e^{st}}{(s-1)(s^2+9)}+\lim_{s\to 3i}(s-3i)\frac{e^{st}}{(s-1)(s^2+9)}\\
&amp;+\lim_{s\to -3i}(s+3i)\frac{e^{st}}{(s-1)(s^2+9)}\\
&amp;= \lim_{s\to 1}\frac{e^{st}}{s^2+9}+\lim_{s\to 3i}\frac{e^{st}}{(s-1)(s+3i)}+\lim_{s\to -3i}\frac{e^{st}}{(s-1)(s-3i)}\\
&amp;\tag{2}
\end{align}
After taking the limit in equation $(2)$ and simplifying, you should get 
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{30}\bigl[3e^t - 3\cos(3t)-\sin(3t)\bigr]
$$