Hello Algebra? It's Me Calculus!

$f(x) = x + \int_{0}^{1} t(x+t)f(t)dt$ $f(x) = x + x\int_{0}^{1} t*f(t)dt + \int_{0}^{1} t^2f(t)dt$

Thus $f(x)$ is a linear function

Let $f(x) = px + q$

$p = 1 + \int_{0}^{1}tf(t)dt$ $q = \int_{0}^{1}t^2f(t)dt$

And,

$tf(t) = pt^2 + qt$ $\therefore p = 1 + \int_{0}^{1}(pt^2 + qt)dt$ $4p = 3q + 6 \cdots (Eqn 1)$

Also,

$q = \int_{0}^{1} (pt^3 + qt^2)dt$ $\therefore \frac{2q}{3} = \frac{p}{4}$ $p = \frac{8q}{3} \cdots (Eqn 2)$

On solving, we get-

$f(x) = \frac{48}{23}x + \frac{18}{23}$ $\therefore \frac{23}{6}f(x) = 8x + 3$

Hence,

$g(x) = a(8x + 3)^2 + b(8x+3) + c$ $g(x) = 64ax^2 + x(8b+48a) + 9a + 3b + c$

The coefficients of $x^2$ , $x$ and $1$ are are equal.

Thus we finally get, $b = 2a$ and $c = 49a$ , and-

$\frac{b+c}{a} - 1 = \frac{49a + 2a}{a} - 1 = 50$