Find a continuous function $f$ that satisfies... Find a continuous function $f$ that satisfies
$$
f(x) = 1 + \frac{1}{x}\int_1^x f(t) \ dt
$$
Note: I tried differentiating with respect to $x$ to get an ODE but you get one that contains integrals - likely difficult to solve.
 A: I will also assume that $f$ is differentiable. We are given that $f(x)=1+\frac{1}{x} \int_1^xf(t)dt$. Multiplying by $x$ we see that
$$xf(x)=x+\int_1^xf(t)dt$$
Differentiating,
$$f(x)+xf'(x)=1+f(x)$$
Subtracting $f(x)$ then dividing through by $x$,
$$f'(x)=\frac{1}{x}$$
Now, integrating we obtain
$$f(x)=\ln(x)+C$$
We must now deal with the initial conditions. from the original condition that $f(1)=1$. So,
$$f(1) = 1 =\ln(1)+C \implies C=1$$
Thus the only continuous (+differentiable) function that satisfies the given condition is 
$$f(x)=\ln(x)+1.$$
A: $\displaystyle F(x) = \int_1^xf(t) dt$
So
$\displaystyle \frac{dF}{dx} = 1 + \frac{1}{x} F$
with condition $F(1)=0$
A: Hint: $xf(x) = x + \int_1^x f(t) \, dt $, except possibly at $x=0$.
Differentiate this to conclude that $f(x) = \ln x + C $. 
Evaluate at $x=1$.
A: Given that
$$f(x)=1+\frac{1}{x}\int_1^x f(t)dt$$ Differentiating both sides
$$ f'(x)=\frac{1}{x}f(x)-\frac{1}{x^2}\int_1^xf(t)dt$$
 $\implies$
$$ f'(x)=\frac{1}{x}f(x)-\frac{1}{x}\left(\frac{1}{x}\int_1^xf(t)dt\right)$$ But
$$\frac{1}{x}\int_1^x f(t)dt=f(x)-1$$ So
$$f'(x)=\frac{f(x}{x}-\frac{1}{x}\left(f(x)-1\right)$$ $\implies$
$$f'(x)=\frac{1}{x} \implies f(x)=Ln(x)+c $$ Finally Use $f(1)=1$ to get the value of $c$
