How do you integrate $((4x^2-1)^2)/x^3dx$ ?

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Answer 1 · 2018-04-12T05:41:26

$8x^2-8 ln x-1/(2x^2)+C$

$((4x^2-1)^2)/x^3 = (16x^4-8x^2+1)/x^3=16x-8/x+1/x^3$

Thus

$int ((4x^2-1)^2)/x^3dx = int (16x-8/x+1/x^3)dx$
$qquad = 8x^2-8 ln x-1/(2x^2)+C$

How do you integrate ((4x^2-1)^2)/x^3dx ((4x^2-1)^2)/x^3dx ?