Question 174085
[b/(b - 1)] + [2b/(b^2 - 1)]

Look for LCD/

The LCD is b^2 - 1.

Divide the LCD by each denominator and then multiply the quotient by each numerator.

b^2 - 1 divided by b - 1 =  b + 1

b^2 - 1 divided by itself = 1

We now have this:

[b(b + 1) + 2b(1)]/(b^2 - 1)

(b^2 + 3b)/(b^2 - 1)

The denominator is the difference of two perfect squares.

So, factor the denominator.

Final answer:

(b^2 + 3b)/(b - 1)(b + 1)