∫ (DX / ((1 + x ^ 1 / 3) x ^ 1 / 2)) to calculate indefinite integral

∫ (DX / ((1 + x ^ 1 / 3) x ^ 1 / 2)) to calculate indefinite integral

∫ 1/[(1+x^1/3)x^1/2] dx
Let x ^ 1 / 6 = u, then x ^ 1 / 2 = u ^ 3, x ^ 1 / 3 = u ^ 2, x = u ^ 6, DX = 6U ^ 5du
=∫ 6u^5/[(1+u^2)u^3] du
=6∫ u^2/(1+u^2) du
=6∫ (u^2+1-1)/(1+u^2) du
=6∫ (u^2+1)/(1+u^2) du - 6∫ 1/(1+u^2) du
=6u - 6arctanu + C
=6x^(1/6) - 6arctan[x^(1/6)] + C
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