$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y)dy$. Find $f(x)$ $f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$. Find $f(x)$ 
I've tried taking $\int_{0}^{1} (xy^2 + x^2y) f(y) dy$ to be $k(x)$ since it comes out to be a function of $x$. That transforms our equation to $f(x) = x + k(x)$.
$f(y) = y + k(y) \implies (xy^2 + x^2y)f(y) = (xy^2 + x^2y)y + (xy^2 + x^2y)k(y)$
Now I tried integrating on both sides against $dy$ from $0$ to $1$ in an attempt to find $k(x)$ 
$\int_{0}^{1} (xy^2 + x^2y)f(y).dy = k(x) = \int_{0}^{1} ((xy^2 + x^2y)y + (xy^2 + x^2y)k(y)).dy$
But I got stuck trying to integrate the right hand side. Any solutions or ideas are appreciated. 

Write you equation under the form :
$$f(x) = x + x \underbrace{\int_{0}^{1} y^2 f(y) dy}_b+ x^2 \underbrace{\int_{0}^{1}y f(y) dy}_a\tag{1}$$
which mean that 
$$f(x)=ax^2+(b+1)x \tag{2}$$ 
with
$$a=\int_{0}^{1}y f(y) dy \ \ (i) \ \ \ \ \text{and} \ \  \ \ b=\int_{0}^{1}y^2 f(y) dy \ \ (ii)\tag{3}$$
is (at most) a second degree polynomial in variable $x$.
It suffices now to plug (2) into (3)(i) and (3)(ii) to get 2 equations in the 2 unknowns $a$ and $b$.
Up to you for the final step.
