Given an integral equation, integrate the function. $$f(x)=x+\int_0^1t(x+t)f(t){\rm d}t$$
Then what is $$\eta=\int_0^1f(x){\rm d}x$$
Ok you can write:
$$\eta=\int_0^1\left(x+\int_0^1t(x+t)f(t)dt\right){\rm d}x=\frac12+\int_0^1\int_0^1t(x+t)f(t){\rm d}t{\rm d}x$$
How to eliminate f?
 A: Reverse the order of integration.
$$\begin{align}\int_0^1 dx \, f(x) &= \frac12 + \int_0^1 dx \ \int_0^1 dt \, t(x+t) f(t)\\ &= \frac12 + \int_0^1 dx \ \int_0^1 dt \, x(t+x) f(x) \\ &= \frac12 + \int_0^1 dx \, x f(x) \left (\frac12+x \right )\\ &= \frac12 + \frac12 \int_0^1 dx \, x f(x) + \int_0^1 dx \, x^2 f(x) \end{align}$$
Note, however, that
$$f(x) = x+ x \int_0^1 dt \, t f(t) + \int_0^1 dt \, t^2 f(t) $$
so that
$$\int_0^1 dx \, f(x) = f \left ( \frac12 \right ) $$
Now, note that
$$f(x) = A x + B$$
where 
$$A = 1+ \int_0^1 dt \, t f(t)$$
$$B = \int_0^1 dt \, t^2 f(t)$$
Then
$$f(x) = x + \int_0^1 dt \, x (x+t) (A t+B) = x + x \left ( \frac13 A + \frac12 B\right ) + \frac14 A + \frac13 B $$
or,
$$A x+B = \left ( 1+\frac13 A + \frac12 B \right ) x + \frac14 A + \frac13 B  $$
so that
$$\frac{2}{3} A - \frac12 B = 1$$
$$\frac14 A - \frac{2}{3} B = 0$$
Then $A=\frac{48}{23}$ and $B = \frac{18}{23}$ and the integral is $f(1/2) = \frac{42}{23}$.
A: Completely expand the RHS and write
$$f(x)=x+\int_0^1t(x+t)f(t){\rm d}t= x +x\int_0^1tf(t){\rm d}t + \int_0^1t^2f(t){\rm d}t =: x+ c_1x+c_2$$
with the constants
$$c_1=\int_0^1tf(t){\rm d}t, \; c_2=\int_0^1t^2f(t){\rm d}t$$
Then
$$\eta=\int_0^1f(x){\rm d}x = \left[\frac{1}{2}(1+c_1)x^2+c_2x\right]_0^1 = \frac{1}{2}(1+c_1)+c_2$$
Edit: To compute $c_1, c_2, \eta$ use
$$c_1=\int_0^1tf(t){\rm d}t = \frac{1}{3}+\frac{1}{3}c_1+\frac{1}{2}c_2$$
$$c_2=\int_0^1t^2f(t){\rm d}t = \frac{1}{4}+\frac{1}{4}c_1+\frac{1}{3}c_2$$
From this linear system you get
$$c_1= \frac{25}{23},\; c_2=\frac{18}{23}$$
and for the integral $\eta=\frac{42}{23}.$
A: This solutions makes use of the Leibniz rule of integral differentiation.
Consider,
\begin{align}
f(x) = x + \int_{0}^{1} t(x+t) f(t) \, dt
\end{align}
for which differentiation leads to
\begin{align}
f'(x) = 1 + \int_{0}^{1} t \, f(t) \, dt.
\end{align}
Differentiate once again to obtain $f''(x) = 0$ which leads to $f(x) = a x + b$. Now, using the intergo-equation it is seen that
\begin{align}
ax + b &= x + \int_{0}^{1} t(x+t) \, (at + b) \, dt \\
&= \left( 1 + \frac{a}{3} + \frac{b}{2} \right) x + \left( \frac{a}{4} + \frac{b}{3} \right)
\end{align}
When the coefficients are equated it is discovered that 
\begin{align}
f(x) = \frac{18}{23} \left( \frac{8 \, x}{3} + 1 \right).
\end{align}
The desired integral to obtain is 
\begin{align}
\int_{0}^{1} f(x) \, dx = \frac{42}{23}.
\end{align}
