Answer
a) $y=\dfrac{x^2}{2}+C$
b) $y=\dfrac{x^3}{3}+C$
c) $y=\dfrac{x^4}{4}+C$
Work Step by Step
a) When $x=(\dfrac{x^2}{2})'$
Thus, $y'=x \implies y=\dfrac{x^2}{2}+C$
b) When $x=(\dfrac{x^3}{3})'$
Thus, $y'=x^2 \implies y=\dfrac{x^3}{3}+C$
c) When $x=(\dfrac{x^4}{4})'$
Thus, $y'=x^3 \implies y=\dfrac{x^4}{4}+C$