Answer
a. $y=\frac{1}{x}+C$.
b. $y=x+\frac{1}{x}+C$.
c. $y=5x-\frac{1}{x}+C$.
Work Step by Step
a. Given $y'=-\frac{1}{x^2}=-x^{-2}$, we can find a function $f(x)=x^{-1}$ such that $f'(x)=-x^{-2}$. Based on Corollary 2, we have $y=f(x)+C=\frac{1}{x}+C$ where $C$ is a constant.
b. Given $y'=1-\frac{1}{x^2}=1-x^{-2}$, we can find a function $f(x)=x+x^{-1}$ such that $f'(x)=1-x^{-2}$. Based on Corollary 2, we have $y=f(x)+C=x+\frac{1}{x}+C$ where $C$ is a constant.
c. Given $y'=5+\frac{1}{x^2}=5+x^{-2}$, we can find a function $f(x)=5x-x^{-1}$ such that $f'(x)=5+x^{-2}$. Based on Corollary 2, we have $y=f(x)+C=5x-\frac{1}{x}+C$ where $C$ is a constant.
