Answer:
[tex]=\frac{3x+2}{x\left(x+1\right)}[/tex]
or
[tex]\frac{(3x+2)}{x^2+x}[/tex]
Step-by-step explanation:
[tex]\frac{2}{x}+\frac{1}{x+1}\\\mathrm{Least\:Common\:Multiplier\:of\:}x,\:x+1:\quad x\left(x+1\right)\\Adjust\:Fractions\:based\:on\:the\:LCM\\Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its\\\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\:x\left(x+1\right)\\\mathrm{For}\:\frac{2}{x}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:x+1\\\frac{2}{x}=\frac{2\left(x+1\right)}{x\left(x+1\right)}\\[/tex]
[tex]\mathrm{For}\:\frac{1}{x+1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:x\\\frac{1}{x+1}=\frac{1\cdot \:x}{\left(x+1\right)x}=\frac{x}{x\left(x+1\right)}\\=\frac{2\left(x+1\right)}{x\left(x+1\right)}+\frac{x}{x\left(x+1\right)}\\\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\=\frac{2\left(x+1\right)+x}{x\left(x+1\right)}\\[/tex]
[tex]\mathrm{Expand}\:2\left(x+1\right)+x:\quad 3x+2\\2\left(x+1\right)+x\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=2,\:b=x,\:c=1\\=2x+2\cdot \:1\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:1=2\\=2x+2\\\mathrm{Simplify}\:2x+2+x:\quad 3x+2\\2x+2+x\\\mathrm{Group\:like\:terms}\\=2x+x+2\\\\mathrm{Add\:similar\:elements:}\:2x+x=3x\\=3x+2\\=\frac{3x+2}{x\left(x+1\right)}[/tex]
