Use integral tables to evaluate the integral. \( \int x \sqrt{19 x-7} d x \)

To evaluate the integral \( \int x \sqrt{19 x - 7} \, dx \), we can use integration by parts or a substitution method. Let's start with a substitution: Let \( u = 19x - 7 \). Then \( du = 19 \, dx \) or \( dx = \frac{du}{19} \). Also, we can express \( x \) in terms of \( u \): \[ x = \frac{u + 7}{19} \] Substituting these into the integral gives: \[ \int x \sqrt{19x - 7} \, dx = \int \left(\frac{u + 7}{19}\right) \sqrt{u} \left(\frac{du}{19}\right) \] \[ = \frac{1}{361} \int (u + 7) \sqrt{u} \, du \] \[ = \frac{1}{361} \left( \int u^{3/2} \, du + 7 \int u^{1/2} \, du \right) \] Now we integrate both parts: 1. For \( \int u^{3/2} \, du \): \[ \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \] 2. For \( \int u^{1/2} \, du \): \[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \] Now putting these results back into our expression: \[ = \frac{1}{361} \left( \frac{2}{5} u^{5/2} + 7 \cdot \frac{2}{3} u^{3/2} \right) + C \] \[ = \frac{2}{1805} u^{5/2} + \frac{14}{361} u^{3/2} + C \] Substituting back \( u = 19x - 7 \): \[ = \frac{2}{1805} (19x - 7)^{5/2} + \frac{14}{361} (19x - 7)^{3/2} + C \] And that gives us the final result for the integral \( \int x \sqrt{19x - 7} \, dx \): \[ \int x \sqrt{19 x - 7} \, dx = \frac{2}{1805} (19x - 7)^{5/2} + \frac{14}{361} (19x - 7)^{3/2} + C \]