Differential Equation with Integration Function $f(x)$ is differentiable. Given further that 
\begin{eqnarray*}
f’(x)+xf’(x-1) &=& 4 \\ \int_0^1 f(xt) dt + \int_0^x f(t-1)dt &=& x^3 +x^2 +2x
\end{eqnarray*}
Determine the function $f(x)$.
$\textbf{Attempt}$
I know $\frac{d}{dx} \int_0^x f(t-1)dt = f(x-1)$, but I am confused about what’s $\frac{d}{dx} \int_0^1 f(xt) dt $ . Maybe I can do $ - \frac{d}{dx} \int_1^x f(xt) dt - \frac{d}{dx} \int_0^x f(xt) dt $ ? 
 A: You can set $xt=u$.
Setting $xt=u$ gives $x=\frac{du}{dt}$, so
$$\int_0^1 f(xt) dt =\int_{0}^{x}f(u)\cdot\frac{du}{x}$$
So, differentiating the both sides of
$$\frac 1x\int_{0}^{x}f(u)du+ \int_0^x f(t-1)dt =x^3 +x^2 +2x$$
gives
$$-\frac{1}{x^2}\int_{0}^{x}f(u)du+\frac{f(x)}{x}+f(x-1)=3x^2+2x+2,$$
i.e.
$$\int_{0}^{x}f(u)du=xf(x)+x^2f(x-1)-3x^4-2x^3-2x^2$$
Now, differentiate the both sides and use $f'(x)+xf'(x-1)=4$.
A: For the proposer's work: 
Using the Leibniz differential rule one has:
$$\frac{d}{dx} \, \int_{a}^{b} f(x, t) \, dt = f(x, b) \frac{d b}{dx} - f(x,a) \frac{d a}{dx} + \int_{a}^{b} \frac{d f}{dx} \, dt,$$ 
where $a = a(x)$, and $b = b(x)$. In this example 
$$\frac{d}{dx} \, \int_{0}^{1} f(x t) \, dt = \int_{0}^{1} t \, f'(x t) \, dt.$$
Since 
$$\int_0^1 f(xt) dt + \int_0^x f(t-1)dt = x^3 +x^2 +2x$$
then differentiation would lead to
$$\int_{0}^{1} t \, f'(x t) \, dt + f(x-1) = 3 x^2 + 2 x + 2.$$
Seemingly this leads to just as much "confusion".
Alternatively one can take the path of the following.
Using 
$$\int_{0}^{1} f(x t) \, dt = \text{ with } u = xt \text{ then } \int_{0}^{x} \frac{f(u) \, du}{x}$$
and
\begin{align}
\int_{0}^{1} f(x t) \, dt + \int_{0}^{x} f(t-1) \, dt &= g(x) = x^3 + x^2 + 2 x \\
\int_{0}^{x} ( f(t) + x \, f(t-1) ) \, dt &= x \, g \\
\frac{d}{dx} \int_{0}^{x} ( f(t) + x \, f(t-1) ) \, dt &= \frac{d}{dx}(x \, g) \\
f(x) + x \, f(x-1) + \int_{0}^{x} f(t-1) \, dt &= \frac{d(x g)}{dx} 
\end{align} 
Differentiation again leads to
\begin{align}
f'(x) + x \, f'(x-1) + f(x-1) + f(x-1) &= \frac{d^2(x g)}{dx^2} \\
4 + 2 \, f(x-1) &= 2(6 x^2 + 3x + 2) \\
f(x-1) &= 6 x^2 + 3 x = 3x \, (2x +1) \\
f(x) &= 6 x^2 + 15 x + 9.
\end{align}
