## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The simplified form of the expression $\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}$ is$\frac{x+3}{{{\left( x+1 \right)}^{2}}}$.
$\frac{2}{{{\left( x+1 \right)}^{2}}}+\frac{1}{x+1}$ 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}