Find the integrable function $g:[0,b]\rightarrow \mathbb R$ in order to satisfy $\int_{0}^{x} t\cdot g(t) dt=x+x^2$ I have found $g(x)=\frac{1}{x}+2$ but it doesn't satisfy for $x=0$ , that is needed.
I also tried to define the function $g(x)=\frac{1}{x}+2$, with $ x>0$ and $ x\leq b$, $g(0)=0$.
Is that possible and if not how do i solve this problem?
Can we use Upper and Lower sums to solve it? (Riemann definition)
 A: If
$$
\int_0^xt\,g(t)\,\mathrm{d}t=x+x^2
$$
Then, for $x\ge0$,
$$
x\int_0^x|g(t)|\,\mathrm{d}t\ge x+x^2
$$
which implies
$$
\lim_{x\to0}\int_0^x|g(t)|\,\mathrm{d}t\ge\lim_{x\to0}\,(1+x)=1
$$
However, if $g(t)$ were integrable, 
$$
\lim_{x\to0}\int_0^x|g(t)|\,\mathrm{d}t=0
$$
Therefore, no integrable $g$ can satisfy the equation.
A: There is no such function. Indeed, suppose it exists. Then
$$\lim_{x \to 0^+} x g(x) = \lim_{x \to 0^+} \frac{1}{x} \int_0^x t \cdot g(t) 
 \ \mathrm{d} t = \lim_{x \to 0^+} 1+x = 1$$
hence $g(x) \sim \frac{1}{x}$ asymptotically as $x \to 0^+$.
This implies that $g$ is not integrable.
A: By differentiating both sides of $\int_{0}^{x} t\cdot g(t) dt=x+x^2$ and applying the Leibniz rule, we deduce that
$$ xg(x) = 1 + 2x$$
Here, let $g(x) = 2 + f(x)$. Then we are left with
$$xf(x) =1$$
Now this has the solution $f(x) = \frac 1x$ which is not good as Crostul pointed out. However, there is a second solution to this equation if you are willing to introduce distributions, namely
$$f(x) = -\delta'(x)$$
Hence $g(t) = 2- \delta'(t)$ is a distributional solution of your equation.
