I have a problem in evaluating the limit as $x$ goes to $0$ of: $$F(x)=\int_1^x f(t)\, dt\,\,\,\ \text{with }f(t)\begin{cases}t^3\ln{t}\, \text{ for }t>0\\ \arctan{t} \, \text{ for } t\leq 0\end{cases}$$ Now I have remarked that since $\lim_{t\to 0^+}f(t)=0=f(0)$ then the function $f$ can be continously extended in $0$ and so the integral function will be continous and defined in $\mathbb{R}$.
If it is continous in $0$ this means that $F(x)\to F(0)=\int_1^0t^3\ln{t}\,dx=-\int_0^1t^3\ln{t}\,dx=\frac{1}{16}$.
$\textbf{Problem}$ $$\lim_{x\to 0^-}\int_{1}^x f(t)=-\lim_{x\to 0^-}\int_{x}^1 f(t)=???$$ In fact $x\to 0^-$ so I have that the extreme $x$ tends to $0^-<0$ but the second is $1$...how can I write this limit and check that it is equal to $\frac{1}{16}$?
If $x < 0$, you have \begin{align*} F(x) &= \int_{1}^{x} f(t)\, dt \\ &= \int_{1}^{0} f(t)\, dt + \int_{0}^{x} f(t)\, dt \\ &= \frac{1}{16} + \int_{0}^{x} \arctan t\, dt. \end{align*}
First problem: Because $F$ is continuous (as are all definite integrals), you know $\lim(F, 0^{-}) = \lim(F, 0^{+}) = \frac{1}{16}$. If you like, this gives you the constant of integration when you antidifferentiate $\arctan$.
Second problem: Your explicit form is not continuous at $0$; the constant in the numerator of the $x \leq 0$ case must agree with $F(0) = \frac{1}{16}$. (And if it matters, you don't need a separate case for $x = 1$.)