\[ \text { Use a substitution of the form } u=a x+b \text { to evaluate the following indefinite integral. } \] \( \int \sqrt{8 x+6} d x \) \( \int \sqrt{8 x+6} d x=\square \)

To evaluate the integral \( \int \sqrt{8x+6} \, dx \) using the substitution \( u = ax + b \), we start by identifying suitable values for \( a \) and \( b \). For this integral, we choose \( u = 8x + 6 \). Thus, the derivative \( du = 8 \, dx \) or \( dx = \frac{du}{8} \). Now, rewrite the integral in terms of \( u \): \[ \int \sqrt{8x+6} \, dx = \int \sqrt{u} \cdot \frac{du}{8} = \frac{1}{8} \int u^{1/2} \, du. \] To evaluate \( \int u^{1/2} \, du \), we use the power rule: \[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}. \] Substituting back, we have: \[ \frac{1}{8} \cdot \frac{2}{3} u^{3/2} = \frac{1}{12} u^{3/2}. \] Now substituting \( u = 8x + 6 \) back into our result gives us: \[ \int \sqrt{8x+6} \, dx = \frac{1}{12} (8x + 6)^{3/2} + C, \] where \( C \) is the constant of integration. To finish up, the final answer is: \[ \int \sqrt{8x+6} \, dx = \frac{1}{12} (8x + 6)^{3/2} + C. \]