Answer
$a=\frac{1}{2}(b^{2}+b)$
Work Step by Step
$\displaystyle \frac{a+1}{b}=\frac{a-1}{b}+\frac{b+1}{a}$
We multiply both sides by $ab$:
$a(a+1)=a(a-1)+b(b+1)$
And distribute:
$a^{2}+a=a^{2}-a+b^{2}+b$
And simplify:
$2a=b^{2}+b$
And solve for $a$:
$a=\frac{1}{2}(b^{2}+b)$
