a) Find the indefinite integral 3x+1x 2 dx . b) The marginal revenue function is given by MR = 3 2x x^2 , x being the output. Find the total revenue function and the demand function.

a) Indefinite Integral: $\int \frac{3x+1}{x-2}dx$

Step 1: Simplify the integrand

The integrand is a rational function $\frac{3x+1}{x-2}$. To simplify, we perform polynomial long division or rewrite the numerator to match the denominator.

We express the numerator $3x+1$ in terms of the denominator $x-2$: $$3x+1=3(x-2)+7$$ This is because: $$3(x-2)=3x-6⟹3x-6+7=3x+1$$

Thus, the integrand becomes: $$\frac{3x+1}{x-2}=\frac{3(x-2)+7}{x-2}=3+\frac{7}{x-2}$$

Step 2: Integrate term by term

Now, the integral can be split into two simpler integrals: $$\int \frac{3x+1}{x-2}dx=\int 3dx+\int \frac{7}{x-2}dx$$

1. The first term is straightforward: $$\int 3dx=3x+C_1$$

2. The second term is a standard logarithmic integral: $$\int \frac{7}{x-2}dx=7\ln |x-2|+C_2$$

Step 3: Combine the results

Adding the results of the two integrals and combining the constants of integration ($C_1+C_2=C$), we get: $$\int \frac{3x+1}{x-2}dx=3x+7\ln |x-2|+C$$

Final Answer: $$\boxed{3x+7\ln |x-2|+C}$$

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b) Marginal Revenue and Demand Function

Given:

The marginal revenue function is: $$MR=3-2x-x^2$$ where $x$ is the output.

Step 1: Find the total revenue function

The total revenue function $TR$ is the integral of the marginal revenue function: $$TR=\int MRdx=\int (3-2x-x^2)dx$$

Integrate term by term: $$\int 3dx=3x$$ $$\int -2xdx=-x^2$$ $$\int -x^{2}dx=-\frac{x^3}{3}$$

Thus, the total revenue function is: $$TR=3x-x^2-\frac{x^3}{3}+C$$

Since revenue cannot be negative and is typically defined up to a constant, we can set $C=0$ for simplicity (assuming $TR=0$ when $x=0$): $$TR=3x-x^2-\frac{x^3}{3}$$

Step 2: Find the demand function

The demand function $D(x)$ is the inverse of the total revenue function in terms of price $P$. However, revenue is defined as: $$TR=P\cdot x$$ where $P$ is the price per unit.

From the total revenue function: $$TR=3x-x^2-\frac{x^3}{3}$$

But $TR=P\cdot x$, so: $$P\cdot x=3x-x^2-\frac{x^3}{3}$$

Divide both sides by $x$ (assuming $x\ne 0$): $$P=3-x-\frac{x^2}{3}$$

Thus, the demand function $D(x)$ is: $$P=3-x-\frac{x^2}{3}$$

Alternatively, if the question expects the demand function in terms of $P$ as a function of $x$, we can write it as: $$D(x)=3-x-\frac{x^2}{3}$$

However, if the demand function is to be expressed as $x$ in terms of $P$, we solve for $x$: $$P=3-x-\frac{x^2}{3}$$ Rearrange: $$\frac{x^2}{3}+x+(P-3)=0$$ Multiply by 3: $$x^2+3x+3(P-3)=0$$

This is a quadratic in $x$: $$x=\frac{-3\pm \sqrt{9-12(P-3)}}{2}$$ $$x=\frac{-3\pm \sqrt{9-12P+36}}{2}$$ $$x=\frac{-3\pm \sqrt{45-12P}}{2}$$

Since $x$ must be non-negative, we take the positive root: $$x=\frac{-3+\sqrt{45-12P}}{2}$$

But this is not typically required unless specified. The more common interpretation is to express $P$ as a function of $x$, which we have already done.

Final Answers:

1. Total Revenue Function: $$\boxed{TR=3x-x^2-\frac{x^3}{3}}$$

2. Demand Function (Price as a function of output): $$\boxed{P=3-x-\frac{x^2}{3}}$$