Question 319682
Also, remember that {{{a^x/a^y=a^(x-y)}}}. So {{{((x^2+19)^21)/((x^2+19)^22)=(x^2+19)^(21-22)=(x^2+19)^(-1)=1/(x^2+19)^1=1/(x^2+19)}}}



Basically, {{{((x^2+19)^21)/((x^2+19)^22)=1/(x^2+19)}}}