1. **State the problem:** We need to find the indefinite integral $$\int (x^3 + e^{3x} + \sin^{-1} x) \, dx$$. 2. **Recall the integral rules:** - The integral of a sum is the sum of the integrals: $$\int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx$$. - Use power rule for polynomials: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$. - For $$\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C$$. - For $$\int \sin^{-1} x \, dx$$, use integration by parts. 3. **Break the integral into parts:** $$\int (x^3 + e^{3x} + \sin^{-1} x) \, dx = \int x^3 \, dx + \int e^{3x} \, dx + \int \sin^{-1} x \, dx$$. 4. **Integrate each part:** - $$\int x^3 \, dx = \frac{x^{4}}{4} + C$$. - $$\int e^{3x} \, dx = \frac{1}{3} e^{3x} + C$$. 5. **Integrate $$\int \sin^{-1} x \, dx$$ using integration by parts:** - Let $$u = \sin^{-1} x$$, so $$du = \frac{1}{\sqrt{1-x^2}} dx$$. - Let $$dv = dx$$, so $$v = x$$. 6. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 7. **Calculate:** $$\int \sin^{-1} x \, dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} \, dx$$. 8. **Evaluate $$\int \frac{x}{\sqrt{1-x^2}} \, dx$$:** - Use substitution $$w = 1 - x^2$$, so $$dw = -2x \, dx$$ or $$-\frac{1}{2} dw = x \, dx$$. - Substitute: $$\int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{x}{\sqrt{w}} \, dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-\frac{1}{2}} dw$$. 9. **Integrate:** $$-\frac{1}{2} \int w^{-\frac{1}{2}} dw = -\frac{1}{2} \cdot 2 w^{\frac{1}{2}} + C = -\sqrt{w} + C = -\sqrt{1-x^2} + C$$. 10. **Substitute back:** $$\int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1-x^2} + C$$. 11. **Combine all parts:** $$\int (x^3 + e^{3x} + \sin^{-1} x) \, dx = \frac{x^{4}}{4} + \frac{1}{3} e^{3x} + x \sin^{-1} x + \sqrt{1-x^2} + C$$. **Final answer:** $$\boxed{\frac{x^{4}}{4} + \frac{1}{3} e^{3x} + x \sin^{-1} x + \sqrt{1-x^2} + C}$$