Explanation

To evaluate the integral \[ \int_1^9 \frac{x(2x-3)}{\sqrt{x}} \, dx, \]

we simplify the integrand: \[ \frac{x(2x-3)}{\sqrt{x}} = 2x^{3/2} - 3x^{1/2}. \]

This allows us to rewrite the integral as: \[ \int_1^9 \left( 2x^{3/2} - 3x^{1/2} \right) \, dx. \] Integrating term by term, we have: \[ \int \left( 2x^{3/2} - 3x^{1/2} \right) \, dx = \frac{4}{5}x^{5/2} - 2x^{3/2} + C. \]

Now we evaluate from 1 to 9: \[ \left[ \frac{4}{5}x^{5/2} - 2x^{3/2} \right]_1^9. \]

Evaluating at \(x = 9\): \[ \frac{4}{5}(9)^{5/2} - 2(9)^{3/2} = \frac{972}{5} - 54 = \frac{702}{5}. \]

Evaluating at \(x = 1\): \[ \frac{4}{5}(1)^{5/2} - 2(1)^{3/2} = \frac{4}{5} - 2 = -\frac{6}{5}. \]

Now, subtract the evaluation at 1 from the evaluation at 9: \[ \frac{702}{5} - \left(-\frac{6}{5}\right) = \frac{708}{5}. \] Thus, the value of the integral is \[ \int_1^9 \frac{x(2x-3)}{\sqrt{x}} \, dx = \frac{708}{5}. \]

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