1. **State the problem:** We need to find the indefinite integral of the function $$3x^3 + 2x^2 e^x - 12x + \frac{5}{x^2}$$ with respect to $$x$$. 2. **Recall the integral rules:** - The integral of $$x^n$$ is $$\frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$. - The integral of $$e^x$$ is $$e^x + C$$. - The integral of a sum is the sum of the integrals. - The integral of $$x^m e^x$$ requires integration by parts, but here we have $$2x^2 e^x$$ which we will integrate by parts. 3. **Break the integral into parts:** $$\int (3x^3) dx + \int (2x^2 e^x) dx - \int (12x) dx + \int \frac{5}{x^2} dx$$ 4. **Integrate each term:** - $$\int 3x^3 dx = 3 \int x^3 dx = 3 \cdot \frac{x^4}{4} = \frac{3x^4}{4}$$ - For $$\int 2x^2 e^x dx$$, use integration by parts: Let $$u = x^2$$, $$dv = 2 e^x dx$$ Then $$du = 2x dx$$, $$v = 2 e^x$$ So, $$\int 2x^2 e^x dx = u v - \int v du = 2x^2 e^x - \int 2 e^x (2x) dx = 2x^2 e^x - 4 \int x e^x dx$$ Now, integrate $$\int x e^x dx$$ by parts again: Let $$u = x$$, $$dv = e^x dx$$ Then $$du = dx$$, $$v = e^x$$ So, $$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C = e^x (x - 1) + C$$ Substitute back: $$\int 2x^2 e^x dx = 2x^2 e^x - 4 e^x (x - 1) + C = e^x (2x^2 - 4x + 4) + C$$ - $$\int (-12x) dx = -12 \int x dx = -12 \cdot \frac{x^2}{2} = -6 x^2$$ - $$\int \frac{5}{x^2} dx = 5 \int x^{-2} dx = 5 \cdot \frac{x^{-1}}{-1} = -\frac{5}{x}$$ 5. **Combine all results:** $$\int (3x^3 + 2x^2 e^x - 12x + \frac{5}{x^2}) dx = \frac{3x^4}{4} + e^x (2x^2 - 4x + 4) - 6 x^2 - \frac{5}{x} + C$$ **Final answer:** $$\boxed{\frac{3x^4}{4} + e^x (2x^2 - 4x + 4) - 6 x^2 - \frac{5}{x} + C}$$