By the way, this is easy, because the antiderivative of $\displaystyle \frac{1}{x^3}=x^{-3}$ is $\displaystyle -\frac{1}{2} x^{-2} +C = \frac{-1}{2x^2} +C$
Determine the antiderivative for $\displaystyle f'(x)=\frac{1-20}{x^3}$,$\displaystyle f(1)=20$
you have $\displaystyle \int{\frac{-19}{x^3}dx}=\frac{19}{2x^2}+C$and you know that $\displaystyle f(1)=20$...so $\displaystyle \frac{19}{2(1)^2}+C=20$...solve for C and you have your original function