Solution of an integral equation $\phi(x)+\int^1_0 xt(x+t)\phi(t)\,dt=x $ , $0 \le x \le 1 $ Solve the following integral equation:  
$\phi(x)+\displaystyle \int^1_0 xt(x+t)\phi(t)\,dt=x $ , $0 \le x \le 1 $

I need to solve the integral equation above. Can anyone help me please?
 A: Hint: $$\phi(x)+x^2 \displaystyle \int_0^1t\phi(t)dt+x\int_0^1t^2\phi(t)=x$$ 
i.e. $$\tag{1}\phi(x)=c_1x^2+c_2x $$
Where $c_1 = - \displaystyle \int_0^1t\phi(t)dt \ \ $ and $\ \ c_2=1- \displaystyle \int_0^1t^2\phi(t)dt$
Put $\phi(t)$ from $(1)$ in these integrals and calculate  $c_1 \ , \ c_2$ from two equations you get 
(two-equations  two variables).
A: Related problems: (I). Here is a detailed solution that maybe someone benefits from it. Rearranging the equation as
$$ \phi(x)= x- \displaystyle \int^1_0 xt(x+t)\phi(t)\,dt \longrightarrow (1). $$
$$  \phi(x) = x - x^2\int_{0}^{1}t\phi(t)\,dt - x \int_{0}^{1}t^2\phi(t) \,dt $$
$$ \implies  \phi(x) = x - x^2 c_1 - x c_2 \longrightarrow (2).$$
Now, just subs back in the integral equation $(1)$ and compare the coefficients of $x's$, you will get a system of 2 equations in $c_1$ and $c_2$
$$ \frac{5}{4}c_{1}+\frac{1}{3}c_{2}= \frac{1}{3} $$
$$ \frac{1}{5}c_{1}+\frac{5}{4}c_{2}= \frac{1}{4}. $$
Solving the above system gives
$$ \left\{ c_{{1}}={\frac {80}{359}},c_{{2}}={\frac {59}{359}} \right\}.  $$
Subs back in $(2)$ yields the solution
$$ \phi \left( x \right) ={\frac {300}{359}}\,x-{\frac {80}{359}}\,{x}^{2
},$$
which can be checked by plugging back in $(1)$. 
