intergrate intergral x^2/sqrt x^6+4 dx.

Question

? +C

Answer 1

Let u = x³. Then du = 3x²dx

So we have (1/3) ∫ [du / (u² + 4) = (1/3) ∫ [du / [u² + 2²] = (1/6)Tan^-1(u/2) + C = (1/6)Tan^-1(x³/2) + C

Answer 2

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