f(x)=\\frac{3}{3x + 2}+3e^{3x}f(0)=4+\\ln 2.\nto advance in the circuit, locate f(1).

Explanation:

Step1: Integrate (f^{\prime}(x))

$$ \begin{align*} f(x)&=\int\left(\frac{3}{3x + 2}+3e^{3x}\right)dx\ &=\int\frac{3}{3x + 2}dx+\int3e^{3x}dx \end{align*} $$ For (\int\frac{3}{3x + 2}dx), let (u = 3x+2), (du=3dx), then (\int\frac{3}{3x + 2}dx=\int\frac{du}{u}=\ln|u|+C_1=\ln|3x + 2|+C_1). For (\int3e^{3x}dx), let (t = 3x), (dt = 3dx), then (\int3e^{3x}dx=\int e^{t}dt=e^{t}+C_2=e^{3x}+C_2). So (f(x)=\ln(3x + 2)+e^{3x}+C) (since (3x+2>0) for the domain where the original function is well - defined, we can drop the absolute value).

Step2: Use the initial condition (f(0)=4+\ln2)

Substitute (x = 0) into (f(x)=\ln(3x + 2)+e^{3x}+C): (f(0)=\ln(3\times0 + 2)+e^{3\times0}+C) (4+\ln2=\ln2 + 1+C) Solve for (C): (C=3)

Step3: Find (f(1))

Substitute (x = 1) into (f(x)=\ln(3x + 2)+e^{3x}+3) (f(1)=\ln(3\times1+2)+e^{3\times1}+3=\ln5+e^{3}+3)

Answer:

(f(1)=\ln5 + e^{3}+3)