1. **Stating the problem:** Evaluate the integral $$\int \frac{3x - \sqrt{Bgsin x}}{\sqrt{1 - x^2}} \, dx$$. 2. **Understanding the integral:** This integral involves a rational expression with a square root in the denominator and a combination of terms in the numerator. We will consider splitting the integral into two parts: $$\int \frac{3x}{\sqrt{1 - x^2}} \, dx - \int \frac{\sqrt{Bgsin x}}{\sqrt{1 - x^2}} \, dx$$ 3. **First integral:** $$I_1 = \int \frac{3x}{\sqrt{1 - x^2}} \, dx$$ Use substitution: let $$u = 1 - x^2$$, then $$du = -2x \, dx$$, so $$x \, dx = -\frac{du}{2}$$. Rewrite: $$I_1 = 3 \int \frac{x}{\sqrt{1 - x^2}} \, dx = 3 \int \frac{x}{\sqrt{u}} \, dx$$ Substitute $$x \, dx = -\frac{du}{2}$$: $$I_1 = 3 \int \frac{-\frac{du}{2}}{\sqrt{u}} = -\frac{3}{2} \int u^{-\frac{1}{2}} \, du$$ Integrate: $$-\frac{3}{2} \cdot 2 u^{\frac{1}{2}} + C = -3 \sqrt{u} + C = -3 \sqrt{1 - x^2} + C$$ 4. **Second integral:** $$I_2 = \int \frac{\sqrt{Bgsin x}}{\sqrt{1 - x^2}} \, dx$$ This integral is complicated due to the presence of $$\sqrt{Bgsin x}$$ and does not simplify easily with elementary functions. Without additional information about the function $$Bgsin x$$ or its relation to $$x$$, this integral cannot be expressed in elementary closed form. 5. **Final answer:** The integral can be expressed as: $$\int \frac{3x - \sqrt{Bgsin x}}{\sqrt{1 - x^2}} \, dx = -3 \sqrt{1 - x^2} - \int \frac{\sqrt{Bgsin x}}{\sqrt{1 - x^2}} \, dx + C$$ where the second integral remains unevaluated without further context. This completes the evaluation of the first part and the setup for the second part.