CUET Mathematics 2026 21st May Shift 2 PYQs

$\dfrac{(x^3-1)^{4/3}}{4} + \dfrac{(x^3-1)^{1/3}}{3} + C$, where $C$ is an arbitrary constant

$\dfrac{(x^3-1)^{7/3}}{7} + \dfrac{(x^3-1)^{4/3}}{4} + C$, where $C$ is an arbitrary constant

$\dfrac{(x^3-1)^{3/7}}{7} + \dfrac{(x^3-1)^{3/4}}{4} + C$, where $C$ is an arbitrary constant

$\dfrac{(x^3-1)^{3/7}}{4} + \dfrac{(x^3-1)^{3/4}}{7} + C$, where $C$ is an arbitrary constant