Answer
The simplified form of the expression $\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}$ is$\frac{x+3}{{{\left( x+1 \right)}^{2}}}$.
Work Step by Step
$\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}$
Solve to obtain the LCD of the denominator of the fractions that is $x+1$and${{\left( x+1 \right)}^{2}}$.
On solving the LCD is${{\left( x+1 \right)}^{2}}$.
The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction:
$\begin{align}
& \frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}=\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}.\frac{x+1}{x+1} \\
& =\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{x+1}{{{\left( x+1 \right)}^{2}}}
\end{align}$
Now, the denominators are same. So, add the numerators and keep the common denominator:
$\begin{align}
& \frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}=\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{x+1}{{{\left( x+1 \right)}^{2}}} \\
& =\frac{2+x+1}{{{\left( x+1 \right)}^{2}}}
\end{align}$
Rearrange the terms and simplify further as follows:
$\begin{align}
& \frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}=\frac{2+x+1}{{{\left( x+1 \right)}^{2}}} \\
& =\frac{x+3}{{{\left( x+1 \right)}^{2}}}
\end{align}$
