How do you integrate $\frac{{(4 x^{2} - 1)}^{2}}{x^{3}} d x$ $((4x^2-1)^2)/x^3dx$ ?

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How do you integrate $\frac{{(4 x^{2} - 1)}^{2}}{x^{3}} d x$ $((4x^2-1)^2)/x^3dx$ ?

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Answer 1 · 2018-04-12T05:41:26

$8 x^{2} - 8 ln x - \frac{1}{2 x^{2}} + C$ $8x^2-8 ln x-1/(2x^2)+C$

Explanation:

$\frac{{(4 x^{2} - 1)}^{2}}{x^{3}} = \frac{16 x^{4} - 8 x^{2} + 1}{x^{3}} = 16 x - \frac{8}{x} + \frac{1}{x^{3}}$ $((4x^2-1)^2)/x^3 = (16x^4-8x^2+1)/x^3=16x-8/x+1/x^3$

Thus

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

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How do you integrate $\frac{{(4 x^{2} - 1)}^{2}}{x^{3}} d x$ $((4x^2-1)^2)/x^3dx$ ?

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Explanation:

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How do you integrate $\frac{{(4 x^{2} - 1)}^{2}}{x^{3}} d x$ $((4x^2-1)^2)/x^3dx$ ?