I need to find a continuous function defined for real and positive $x$ such that $f(x) = 1 + \frac1{x} \int_1^x f(t)\ \mathrm{d}t$. What I did is the following:
$$\begin{align*}f(x) &= 1 + \frac1{x} \int_1^x f(t)\ \mathrm{d}t\\ x(f(x) - 1) &= \int_1^x f(t) \ \mathrm{d}t \end{align*}$$
Differentiating both sides:
$$\begin{align*} f(x) - 1 + x f'(x) &= f(x)\\ f'(x) &= \frac1{x}\\ f(x) &= \ln x \end{align*}$$
Which looks alright. But when checking to make sure, I get this:
$$\begin{align*} f(x) &= 1 + \frac1{x} \int_1^x \ln t \ \mathrm{d}t\\ f(x) &= 1 + \frac1{x} (x(\ln x -1)|_1^x)\\ f(x) &= 1 + \frac1{x} (x(\ln x -1) - (\ln 1 - 1))\\ f(x) &= 1 + \ln x - 1 + \frac1{x}\\ f(x) &= \ln x + \frac1{x} \end{align*}$$
which is different from what I got before. What is my mistake here?
You left out the constant of integration... the answer is actually $\ln(x) + 1$.