1. The problem is to find the indefinite integral of the function $$f(x) = \frac{2}{\sqrt{x}} + 6x - 3 \cdot 2^{4x} + \left(\frac{1}{3}\right)^{-x}$$ with respect to $x$. 2. Recall the integral formulas: - $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ for $n \neq -1$. - $$\int a^{x} \, dx = \frac{a^{x}}{\ln(a)} + C$$ for $a > 0$, $a \neq 1$. 3. Rewrite terms for easier integration: - $$\frac{2}{\sqrt{x}} = 2x^{-\frac{1}{2}}$$ - $$\left(\frac{1}{3}\right)^{-x} = 3^{x}$$ because $$\left(\frac{1}{3}\right)^{-x} = 3^{x}$$. 4. Integrate each term separately: - $$\int 2x^{-\frac{1}{2}} \, dx = 2 \int x^{-\frac{1}{2}} \, dx = 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2 \cdot 2x^{\frac{1}{2}} + C = 4\sqrt{x} + C$$ - $$\int 6x \, dx = 6 \cdot \frac{x^{2}}{2} + C = 3x^{2} + C$$ - $$\int -3 \cdot 2^{4x} \, dx = -3 \int 2^{4x} \, dx$$ Use substitution for $$\int 2^{4x} dx$$: Let $$u = 4x$$, then $$du = 4 dx$$ or $$dx = \frac{du}{4}$$. So, $$\int 2^{4x} dx = \int 2^{u} \frac{du}{4} = \frac{1}{4} \int 2^{u} du = \frac{1}{4} \cdot \frac{2^{u}}{\ln(2)} + C = \frac{2^{4x}}{4 \ln(2)} + C$$ Therefore, $$\int -3 \cdot 2^{4x} dx = -3 \cdot \frac{2^{4x}}{4 \ln(2)} + C = -\frac{3}{4 \ln(2)} 2^{4x} + C$$ - $$\int 3^{x} \, dx = \frac{3^{x}}{\ln(3)} + C$$ 5. Combine all results: $$\int f(x) dx = 4\sqrt{x} + 3x^{2} - \frac{3}{4 \ln(2)} 2^{4x} + \frac{3^{x}}{\ln(3)} + C$$ This is the indefinite integral of the given function.