Answer
$\dfrac{-b^2+b+2}{-b^2+b}$
Work Step by Step
Simplify by multiplying the LCD (which is $(1-b)(1+b)$) to both the numerator and the denominator to obtain:
$\require{cancel}
=\dfrac{(1-b)(1+b)\left(1+\frac{1}{1-b}\right)}{(1-b)(1+b)(1-\frac{1}{1+b})}
\\=\dfrac{(1-b)(1+b)(1)+(1-b)(1+b)(\frac{1}{1-b})}{(1-b)(1+b)(1)-(1-b)(1+b)(\frac{1}{1+b})}
\\=\dfrac{(1-b)(1+b)+(1+b)}{(1-b)(1+b)-(1-b)}$
Use the formula $(x-y)(x+y)=x^2-y^2$ to obtain:
$=\dfrac{(1^2-b^2)+1+b}{(1^2-b^2)-1+b}
\\=\dfrac{1-b^2+1+b}{1-b^2-1+b}
\\=\dfrac{2+b-b^2}{b-b^2}
\\=\dfrac{-b^2+b+2}{-b^2+b}$
