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

1 Answer

Ananda Dasgupta

Apr 12, 2018

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

Explanation:

#((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#

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