The integral (x^8 - x^2)dx (x^ 12 + 3x^6 + 1) ^ -1 (x^3 + 1 x^3 ) is equal to:

JEE Main 2024 Mathematics question

The integral is equal to:

Answer: (C)

Let t = tan⁻¹(x³ + 1/x³). Then dt = [1/(1 + (x³ + 1/x³)²)]·3(x² − 1/x⁴)dx = 3(x⁸ −

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Question ID VK-016628