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

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To integrate (\int \frac{x^2 - 1}{x^{3/2}} , dx), use the method of integration by parts. First, rewrite the integral as (\int x^{2 - 3/2} - x^{-1/2} , dx). Then, apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Let ( u = x^{2 - 3/2} ) and ( dv = dx ), so ( du = (2 - 3/2)x^{2 - 3/2 - 1} , dx = (2 - 3/2)x^{1/2} , dx ) and ( v = x ).

Substitute these values into the integration by parts formula:

[ \int x^{2 - 3/2} , dx - \int x^{-1/2} , dx = \frac{x^{7/2}}{7/2} - \frac{2x^{1/2}}{1/2} + C ]

[ = \frac{2}{7}x^{7/2} - 4x^{1/2} + C ]

Therefore, the integral of ( \frac{x^2 - 1}{x^{3/2}} , dx ) is ( \frac{2}{7}x^{7/2} - 4x^{1/2} + C ), where ( C ) is the constant of integration.

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