Answer
$10\sqrt{3}$
Work Step by Step
\begin{aligned}
\int_{-\sqrt{3}}^{\sqrt{3}}(t+1)\left(t^{2}+4\right) d t&= \int _{-\sqrt{3}}^{\sqrt{3}}(t^3+4t+t^2+4)dt\\
&=\int _{-\sqrt{3}}^{\sqrt{3}}t^3dt+\int _{-\sqrt{3}}^{\sqrt{3}}4tdt+\int _{-\sqrt{3}}^{\sqrt{3}}t^2dt+\int _{-\sqrt{3}}^{\sqrt{3}}4dt
\end{aligned}
Since $t^3$ and $4t$ are odd functions , then
\begin{aligned}
\int _{-\sqrt{3}}^{\sqrt{3}}t^3dt&=0\\
\int _{-\sqrt{3}}^{\sqrt{3}}4tdt&=0
\end{aligned}
Hence
\begin{aligned}
\int_{-\sqrt{3}}^{\sqrt{3}}(t+1)\left(t^{2}+4\right) d t&= \int _{-\sqrt{3}}^{\sqrt{3}}(t^3+4t+t^2+4)dt\\
&=\int _{-\sqrt{3}}^{\sqrt{3}}t^3dt+\int _{-\sqrt{3}}^{\sqrt{3}}4tdt+\int _{-\sqrt{3}}^{\sqrt{3}}t^2dt+\int _{-\sqrt{3}}^{\sqrt{3}}4dt\\
&= 0+0+\frac{1}{3}t^3+4t\bigg|_{-\sqrt{3}}^{\sqrt{3}}\\
&=0+0+2\sqrt{3}+8\sqrt{3}\\
&=10\sqrt{3}
\end{aligned}
