Question 153182
{{{b/(b-1)+(2b)/(b^2-1)}}} Start with the given expression.



{{{b/(b-1)+(2b)/((b-1)(b+1))}}} Factor {{{b^2-1}}} to get {{{(b-1)(b+1)}}}. Notice how the LCD is {{{(b-1)(b+1)}}}



{{{((b+1)/(b+1))(b/(b-1))+(2b)/((b-1)(b+1))}}} Multiply the first term by {{{(b+1)/(b+1)}}}. The goal is to get the first denominator equal to the LCD {{{(b-1)(b+1)}}}



{{{(b(b+1))/((b-1)(b+1))+(2b)/((b-1)(b+1))}}} Combine the fractions.



{{{(b^2+b)/((b-1)(b+1))+(2b)/((b-1)(b+1))}}} Distribute



{{{(b^2+b+2b)/((b-1)(b+1))}}} Add the fractions.



{{{(b^2+3b)/((b-1)(b+1))}}} Combine like terms.



So {{{b/(b-1)+(2b)/(b^2-1)=(b^2+3b)/((b-1)(b+1))}}} where {{{b<>-1}}} or {{{b<>1}}}