Answer
$\frac{5a-3}{a\left( a+1 \right)\left( a-1 \right)}$
Work Step by Step
$\begin{align}
& \frac{3}{{{a}^{2}}+a}+\frac{4}{2{{a}^{2}}-2}=\frac{3}{a\left( a+1 \right)}+\frac{4}{2\left( {{a}^{2}}-{{1}^{2}} \right)} \\
& =\frac{3}{a\left( a+1 \right)}+\frac{2}{\left( a-1 \right)\left( a+1 \right)} \\
& =\frac{1}{a+1}\left( \frac{3}{a}+\frac{2}{a-1} \right)
\end{align}$
And,
$\begin{align}
& \frac{1}{a+1}\left( \frac{3}{a}+\frac{2}{a-1} \right)=\frac{1}{a+1}\left( \frac{3\times \left( a-1 \right)+2\times a}{a\left( a-1 \right)} \right) \\
& =\frac{3a-3+2a}{a\left( a-1 \right)\left( a+1 \right)} \\
& =\frac{5a-3}{a\left( a-1 \right)\left( a+1 \right)}
\end{align}$
Thus,
$\frac{3}{{{a}^{2}}+a}+\frac{4}{2{{a}^{2}}-2}=\frac{5a-3}{a\left( a+1 \right)\left( a-1 \right)}$
Thus, the simplified form of the expression $\frac{3}{{{a}^{2}}+a}+\frac{4}{2{{a}^{2}}-2}$ is $\frac{5a-3}{a\left( a+1 \right)\left( a-1 \right)}$.
