9. Evaluate the indefinite integral. (Use \( C \) for the constant of integration.) \[ \int \frac{a+b x^{5}}{\sqrt{6 a x+b x^{6}}} d x \]

To evaluate the indefinite integral \[ \int \frac{a + b x^{5}}{\sqrt{6 a x + b x^{6}}} \, dx, \] let's use a substitution. Set \[ u = 6 a x + b x^{6}. \] Then compute the derivative of \( u \): \[ \frac{du}{dx} = 6a + 6bx^{5}, \] which simplifies to \[ du = (6a + 6bx^{5}) dx = 6(a + bx^{5}) \, dx. \] Now we can express \( dx \): \[ dx = \frac{du}{6(a + bx^{5})}. \] Substituting in our integral, we have: \[ \int \frac{a + b x^{5}}{\sqrt{u}} \cdot \frac{du}{6(a + bx^{5})}. \] Notice that the \( (a + bx^{5}) \) terms cancel, leading to: \[ \frac{1}{6} \int \frac{1}{\sqrt{u}} \, du. \] Now, we can easily integrate this: \[ \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C. \] Substituting back \( u = 6 a x + b x^{6} \), we find: \[ \frac{1}{6} \cdot 2\sqrt{6 a x + b x^{6}} + C = \frac{1}{3}\sqrt{6 a x + b x^{6}} + C. \] Thus, the final result for the indefinite integral is: \[ \int \frac{a + b x^{5}}{\sqrt{6 a x + b x^{6}}} \, dx = \frac{1}{3}\sqrt{6 a x + b x^{6}} + C. \]