Answer
The simplified form of the expression $\frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}$ is $\frac{{{x}^{2}}+2x+{{y}^{2}}+2y}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}$.
Work Step by Step
$\frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}$
Solve to obtain the LCD of denominator of the fractions that is $\left( x+y \right)$ and $\left( {{x}^{2}}+{{y}^{2}} \right)$.
On solving the LCD is $\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)$.
The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction:
$\begin{align}
& \frac{1}{x+y}=\frac{1\left( {{x}^{2}}+{{y}^{2}} \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \\
& =\frac{\left( {{x}^{2}}+{{y}^{2}} \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}
\end{align}$
And
$\frac{2}{{{x}^{2}}+{{y}^{2}}}=\frac{2\left( x+y \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}$
Since the fractions have the common denominator, add the numerators and the LCD is the denominator:
$\begin{align}
& \frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}=\frac{\left( {{x}^{2}}+{{y}^{2}} \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}+\frac{2\left( x+y \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \\
& =\frac{\left( {{x}^{2}}+{{y}^{2}} \right)+2\left( x+y \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}
\end{align}$
Simplify the above expression as:
$\begin{align}
& \frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}=\frac{{{x}^{2}}+{{y}^{2}}+2x+2y}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \\
& =\frac{{{x}^{2}}+2x+{{y}^{2}}+2y}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}
\end{align}$