x^8(x^3+1)^1/3 dx is equal to Option 1: 1/4(x^3+1)^4/3 - 1/2(x^3+1)^2/3 - 2/5(x^3+1)^1/3 + C, C is an arbitrary constant Option 2: 1/4(x^3+1)^4/3 + 1/2(x^3+1)^2/3 - 2/5(x^3+1)^1/3 + C, C is an arbitrary constant Option 3: 1/8(x^3+1)^8/3 - 1/2(x^3+1)^2/3 + 2/5(x^3+1)^5/3 + C, C is an arbitrary constant Option 4: 1/8(x^3+1)^8/3 + 1/2(x^3+1)^2/3 - 2/5(x^3+1)^5/3 + C, C is an arbitrary constant

The answer is '1/8(x^3+1)^8/3 + 1/2(x^3+1)^2/3 - 2/5(x^3+1)^5/3 + C, C is an arbitrary constant'. View full solution:

CUET Mathematics - x^8(x^3+1)^1/3 dx is equal to | PYQs + Solutions

CUET Mathematics 2026 14th May Shift 2

Calculus

Medium

common

$\frac{1}{4}(x^3+1)^{4/3} - \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{1/3} + C$ , C is an arbitrary constant

$\frac{1}{4}(x^3+1)^{4/3} + \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{1/3} + C$ , C is an arbitrary constant

$\frac{1}{8}(x^3+1)^{8/3} - \frac{1}{2}(x^3+1)^{2/3} + \frac{2}{5}(x^3+1)^{5/3} + C$ , C is an arbitrary constant

$\frac{1}{8}(x^3+1)^{8/3} + \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{5/3} + C$ , C is an arbitrary constant

✅ Correct Option: 4