#### Answer

$\frac{2x+1}{x^{2}(x+1)}$

#### Work Step by Step

$\frac{1}{x^{2}}+\frac{1}{x^{2}+x}$
Factor out the common term from the second fraction:
$=\frac{1}{x^{2}}+\frac{1}{x(x+1)}$
Find the least common denominator (i.e. $x^{2}(x+1)$) and adjust the fractions accordingly:
$=\frac{1\times (x+1)}{x^{2}(x+1)}+\frac{1\times x}{x^{2}(x+1)}$
Simplify the fractions:
$=\frac{x+1}{x^{2}(x+1)}+\frac{x}{x^{2}(x+1)}$
Combine the fractions:
$=\frac{x+1+x}{x^{2}(x+1)}$
Collect the like terms:
$=\frac{2x+1}{x^{2}(x+1)}$