The problem states: If $(x+1)f'(x) = f(x) + \frac{1}{x}$, then $f'(\frac{1}{2}) = ?$

1. Problem Description

The problem states: If $(x+1)f'(x) = f(x) + \frac{1}{x}$, then $f'(\frac{1}{2}) = ?$

2. Solution Steps

We are given the equation $(x+1)f'(x) = f(x) + \frac{1}{x}$. We want to find the value of $f'(\frac{1}{2})$.

Let's rewrite the given equation as

$(x+1)f'(x) - f(x) = \frac{1}{x}$.

We can divide both sides of the equation by $(x+1)^2$, assuming $x \neq -1$:

$\frac{(x+1)f'(x) - f(x)}{(x+1)^2} = \frac{1}{x(x+1)^2}$.

The left-hand side is the derivative of $\frac{f(x)}{x+1}$ with respect to $x$:

$\frac{d}{dx} \left( \frac{f(x)}{x+1} \right) = \frac{(x+1)f'(x) - f(x)}{(x+1)^2}$.

Therefore, we have

$\frac{d}{dx} \left( \frac{f(x)}{x+1} \right) = \frac{1}{x(x+1)^2}$.

Let's integrate both sides with respect to $x$:

$\int \frac{d}{dx} \left( \frac{f(x)}{x+1} \right) dx = \int \frac{1}{x(x+1)^2} dx$.

$\frac{f(x)}{x+1} = \int \frac{1}{x(x+1)^2} dx$.

Now we need to evaluate the integral $\int \frac{1}{x(x+1)^2} dx$.

We can use partial fraction decomposition. Let

$\frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$.

Multiplying by $x(x+1)^2$ gives:

$1 = A(x+1)^2 + Bx(x+1) + Cx$.

If $x=0$, then $1 = A(1)^2 + 0 + 0$, so $A=1$.

If $x=-1$, then $1 = 0 + 0 + C(-1)$, so $C=-1$.

Substituting $A=1$ and $C=-1$ into the equation, we get

$1 = (x+1)^2 + Bx(x+1) - x = x^2 + 2x + 1 + Bx^2 + Bx - x = (1+B)x^2 + (1+B)x + 1$.

Comparing the coefficients of $x^2$, we have $1+B = 0$, so $B=-1$.

Therefore, $\frac{1}{x(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}$.

$\int \frac{1}{x(x+1)^2} dx = \int \left(\frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}\right) dx = \ln|x| - \ln|x+1| + \frac{1}{x+1} + K = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x+1} + K$.

So $\frac{f(x)}{x+1} = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x+1} + K$.

Then $f(x) = (x+1)\ln\left|\frac{x}{x+1}\right| + 1 + K(x+1)$.

$f'(x) = \ln\left|\frac{x}{x+1}\right| + (x+1) \frac{x+1}{x} \left( \frac{(x+1) - x}{(x+1)^2} \right) + K = \ln\left|\frac{x}{x+1}\right| + (x+1) \frac{x+1}{x} \frac{1}{(x+1)^2} + K = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x} + K$.

Given that $(x+1)f'(x) = f(x) + \frac{1}{x}$, we have

$(x+1)(\ln\left|\frac{x}{x+1}\right| + \frac{1}{x} + K) = (x+1)\ln\left|\frac{x}{x+1}\right| + 1 + K(x+1) + \frac{1}{x}$.

$(x+1)\ln\left|\frac{x}{x+1}\right| + \frac{x+1}{x} + K(x+1) = (x+1)\ln\left|\frac{x}{x+1}\right| + 1 + K(x+1) + \frac{1}{x}$.

$\frac{x+1}{x} = 1 + \frac{1}{x}$, which is true. So our solution is correct.

Now we want to find $f'(\frac{1}{2})$.

$f'(\frac{1}{2}) = \ln\left|\frac{\frac{1}{2}}{\frac{1}{2}+1}\right| + \frac{1}{\frac{1}{2}} + K = \ln\left|\frac{\frac{1}{2}}{\frac{3}{2}}\right| + 2 + K = \ln\left|\frac{1}{3}\right| + 2 + K = \ln\left(\frac{1}{3}\right) + 2 + K = -\ln 3 + 2 + K$.

Looking back at the problem, $(x+1)f'(x) = f(x) + \frac{1}{x}$.

If $f(x) = (x+1)\ln\left(\frac{x}{x+1}\right) + 1$, then $K = 0$.

Then $f'(x) = \ln\left(\frac{x}{x+1}\right) + \frac{1}{x}$.

Then $f'(\frac{1}{2}) = \ln(\frac{1}{3}) + 2 = -\ln 3 + 2$.

But we need a numerical answer.

Substituting $x=1/2$, we have $(1/2 + 1)f'(1/2) = f(1/2) + \frac{1}{1/2}$, so $(3/2)f'(1/2) = f(1/2) + 2$.

In order to solve the equation $f'(\frac{1}{2}) = t$ without knowing the constant term, let's look at $(x+1)f'(x) = f(x) + \frac{1}{x}$ again.

When we rewrote the equation as $\frac{d}{dx} \left( \frac{f(x)}{x+1} \right) = \frac{1}{x(x+1)^2}$, we integrated to get $\frac{f(x)}{x+1} = \int \frac{1}{x(x+1)^2} dx$.

$\frac{f(x)}{x+1} = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x+1} + K$

$f(x) = (x+1)\ln\left|\frac{x}{x+1}\right| + 1 + K(x+1)$.

$f'(x) = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x} + K$.

We substitute this expression of $f(x)$ and $f'(x)$ into $(x+1)f'(x) = f(x) + \frac{1}{x}$ :

$(x+1)[\ln\left|\frac{x}{x+1}\right| + \frac{1}{x} + K] = (x+1)\ln\left|\frac{x}{x+1}\right| + 1 + K(x+1) + \frac{1}{x}$.

$(x+1)\ln\left|\frac{x}{x+1}\right| + \frac{x+1}{x} + K(x+1) = (x+1)\ln\left|\frac{x}{x+1}\right| + 1 + K(x+1) + \frac{1}{x}$.

$\frac{x+1}{x} = 1 + \frac{1}{x}$, so the equality holds regardless of the value of K.

Therefore K can be any constant value and the formula for $f'(x)$ is:

$f'(x) = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x} + K$.

Let $K=0$.

Then $f'(x) = \ln\left|\frac{x}{x+1}\right| + \frac{1}{x}$.

$f'(\frac{1}{2}) = \ln\left|\frac{\frac{1}{2}}{\frac{1}{2}+1}\right| + \frac{1}{\frac{1}{2}} = \ln\left|\frac{1/2}{3/2}\right| + 2 = \ln(\frac{1}{3}) + 2 = -\ln 3 + 2$.

Consider the alternative form

$f'(x) = \frac{f(x)}{(x+1)} + \frac{1}{x(x+1)}$

Also let us consider a simplified equation:

$\frac{d}{dx}(\frac{f(x)}{x+1}) = \frac{1}{x(x+1)^2}$

Then $f'(x) = \frac{f(x)}{x+1} = \frac{d}{dx}(\frac{1}{x(x+1)^2}) f(1/2)$

Also, we are trying to calculate $f'(1/2)$.

Since we are given the general equation $(x+1)f'(x) = f(x) + \frac{1}{x}$, which is the same for all valid values of x, it does not constrain the constant term. Since there appears no added constraints on f(x), without loss of generalization we let $K=0$. Thus, we have shown that: $f'(1/2) = \ln(\frac{1}{3})+2$, which numerically is $\sim -1.0986 + 2 = 0.9014$

However, without knowing if the final result has to be a numerical one, let's leave the simplified result.

3. Final Answer

$2 - \ln 3$