1. **State the problem:** We need to find the integral $$\int \left(x^2 + e^{3x} + \sin^{-1}(x)\right) dx$$. 2. **Split the integral:** Using the linearity of integration, we write: $$\int x^2 dx + \int e^{3x} dx + \int \sin^{-1}(x) dx$$ 3. **Integrate each term separately:** - For $$\int x^2 dx$$, use the power rule: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ So, $$\int x^2 dx = \frac{x^{3}}{3} + C_1$$ - For $$\int e^{3x} dx$$, use substitution or recall that: $$\int e^{ax} dx = \frac{1}{a} e^{ax} + C$$ So, $$\int e^{3x} dx = \frac{1}{3} e^{3x} + C_2$$ - For $$\int \sin^{-1}(x) dx$$, use integration by parts: Let $$u = \sin^{-1}(x)$$ and $$dv = dx$$. Then, $$du = \frac{1}{\sqrt{1 - x^2}} dx$$ and $$v = x$$. Integration by parts formula: $$\int u dv = uv - \int v du$$ Apply: $$\int \sin^{-1}(x) dx = x \sin^{-1}(x) - \int \frac{x}{\sqrt{1 - x^2}} dx$$ 4. **Evaluate the remaining integral:** $$\int \frac{x}{\sqrt{1 - x^2}} dx$$ Use substitution: Let $$w = 1 - x^2$$, so $$dw = -2x dx$$ or $$-\frac{1}{2} dw = x dx$$. Rewrite integral: $$\int \frac{x}{\sqrt{1 - x^2}} dx = \int \frac{x}{\sqrt{w}} dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-1/2} dw$$ Integrate: $$-\frac{1}{2} \cdot 2 w^{1/2} + C = -\sqrt{w} + C = -\sqrt{1 - x^2} + C$$ 5. **Combine results:** $$\int \sin^{-1}(x) dx = x \sin^{-1}(x) + \sqrt{1 - x^2} + C_3$$ 6. **Write the full integral:** $$\int \left(x^2 + e^{3x} + \sin^{-1}(x)\right) dx = \frac{x^3}{3} + \frac{1}{3} e^{3x} + x \sin^{-1}(x) + \sqrt{1 - x^2} + C$$ where $$C = C_1 + C_2 + C_3$$ is the constant of integration. This completes the integration step-by-step.