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Verified answer
2 integral 1 [(3x^4 - 4x^3 + 2)/x^2] dx= 2 integral 1 [3x^2 - 4x + 2x^-2] dx
......................................2
= [x^3 - 2x^2 - 2x^-1]
......................................1
= [2^3 - 2(2^2) - 2(2^-1)] - [1^3 - 2(1^2) - 2(1^-1)]
= (8 - 8 - 1) - (1 - 2 - 2)
= -1 - (-3)
= -1 + 3
= 2