Integrate this by part -x∫eˣdx + 1/x∫x²eˣdx

The integral of $e^x$ is straightforward:

$$\int e^x dx = e^x + C_1$$

The integration by parts formula is $\int u \, dv = uv - \int v \, du$.
For $\int x^2 e^x dx$:
Let $u = x^2$, then $du = 2x \, dx$.
Let $dv = e^x dx$, then $v = e^x$.
Applying the formula:

$$\int x^2 e^x dx = x^2 e^x - \int e^x (2x) dx = x^2 e^x - 2 \int x e^x dx$$

For $\int x e^x dx$:
Let $u = x$, then $du = dx$.
Let $dv = e^x dx$, then $v = e^x$.
Applying the formula:

$$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C_2$$

$$\int x^2 e^x dx = x^2 e^x - 2 (x e^x - e^x + C_2)$$ $$= x^2 e^x - 2x e^x + 2e^x - 2C_2$$ $$= e^x(x^2 - 2x + 2) + C_3 \quad (\text{where } C_3 = -2C_2)$$

$$-x \int e^x dx + \frac{1}{x} \int x^2 e^x dx$$ $$= -x (e^x + C_1) + \frac{1}{x} (e^x(x^2 - 2x + 2) + C_3)$$ $$= -x e^x - x C_1 + \frac{e^x}{x}(x^2 - 2x + 2) + \frac{C_3}{x}$$ $$= -x e^x - x C_1 + e^x \left( \frac{x^2}{x} - \frac{2x}{x} + \frac{2}{x} \right) + \frac{C_3}{x}$$ $$= -x e^x - x C_1 + e^x \left( x - 2 + \frac{2}{x} \right) + \frac{C_3}{x}$$ $$= -x e^x - x C_1 + x e^x - 2e^x + \frac{2e^x}{x} + \frac{C_3}{x}$$

$$= (-x e^x + x e^x) - 2e^x + \frac{2e^x}{x} - x C_1 + \frac{C_3}{x}$$ $$= -2e^x + \frac{2e^x}{x} - x C_1 + \frac{C_3}{x}$$ $$= 2e^x \left( \frac{1}{x} - 1 \right) + C$$

(where $C$ represents the combined arbitrary constants $-x C_1 + \frac{C_3}{x}$)

The simplified expression is:

$$2e^x \left( \frac{1}{x} - 1 \right) + C$$