1. **State the problem:** We need to evaluate the integral $$\int \frac{5xe^{4x} - 2e^x}{10xe^x} \, dx$$ 2. **Simplify the integrand:** Divide numerator and denominator term-by-term: $$\frac{5xe^{4x}}{10xe^x} - \frac{2e^x}{10xe^x} = \frac{5xe^{4x}}{10xe^x} - \frac{2e^x}{10xe^x}$$ 3. Cancel common factors in each term: For the first term: $$\frac{5\cancel{x}e^{4x}}{10\cancel{x}e^x} = \frac{5e^{4x}}{10e^x}$$ For the second term: $$\frac{2e^x}{10x e^x} = \frac{2\cancel{e^x}}{10x\cancel{e^x}} = \frac{2}{10x}$$ 4. Simplify further: $$\frac{5e^{4x}}{10e^x} = \frac{5}{10} e^{4x - x} = \frac{1}{2} e^{3x}$$ 5. So the integrand becomes: $$\frac{1}{2} e^{3x} - \frac{1}{5x}$$ 6. Rewrite the integral: $$\int \left( \frac{1}{2} e^{3x} - \frac{1}{5x} \right) dx = \frac{1}{2} \int e^{3x} dx - \frac{1}{5} \int \frac{1}{x} dx$$ 7. Integrate each term: - For $$\int e^{3x} dx$$, use substitution: Let $$u = 3x$$, then $$du = 3 dx$$, so $$dx = \frac{du}{3}$$. Therefore, $$\int e^{3x} dx = \int e^u \frac{du}{3} = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x} + C$$ - For $$\int \frac{1}{x} dx$$, the integral is: $$\ln|x| + C$$ 8. Substitute back into the integral: $$\frac{1}{2} \times \frac{1}{3} e^{3x} - \frac{1}{5} \ln|x| + C = \frac{1}{6} e^{3x} - \frac{1}{5} \ln|x| + C$$ **Final answer:** $$\int \frac{5xe^{4x} - 2e^x}{10xe^x} \, dx = \frac{1}{6} e^{3x} - \frac{1}{5} \ln|x| + C$$