It is possible to integrate this function? I  heard that  that there are some unintegratable functions and I want to as if this one is not one of them?
\begin{equation}
\large \int \frac{t}{t+1}dt
\end{equation}
Got this by trying to solve another function and I need to check if this is not a dead end. If it is I will just have to find another way.  It is likely that I just lack some skill and knowledge  to solve it despite it`s simple appearance.

This integral is easy to do:
$$\begin{align*}
\int \frac{t}{t+1}\,dt &= \int\frac{t+1-1}{t+1}\,dt\\
 &= \int\left(\frac{t+1}{t+1}-\frac{1}{t+1}\right)\,dt\\
&= \int1\,dt - \int\frac{dt}{t+1}\\
&= t - \ln|t+1| + C.
\end{align*}$$
You can verify this by differentiation:
$$\frac{d}{dt}\left(t - \ln|t+1| + C\right) = 1 - \frac{1}{t+1} = \frac{t+1-1}{t+1} = \frac{t}{t+1}.$$
Note. What you have here is the integral of a rational function (a polynomial divided by a polynomial). In principle, every rational function has an elementary integral. There's even an algorithm for finding them.
To find the integral of $\frac{p(t)}{q(t)}$, where $p$ and $q$ are polynomials:


*

*If $\deg(p)\geq \deg(q)$, then perform long division with remainder and rewrite the fraction as
$$\frac{p(t)}{q(t)} = P(t) + \frac{r(t)}{q(t)}$$
where $P(t)$ is a polynomial, and $r(t)$ is a polynomial with $r=0$ or $\deg(r)\lt\deg(q)$. $P(t)$ can be integrated easily, so we are left with the problem of integrating rational functions $\frac{p(t)}{q(t)}$ with $\deg(p)\lt\deg(q)$.

*Completely factor $q(t)$ into a product of linear and irreducible quadratic polynomials. This step can be hard to perform in practice! In fact, this is the only reason why I say "in principle" above, because actually factoring a polynomial can be very hard to do.

*Use Partial Fraction Decomposition to rewrite $\frac{p(t)}{q(t)}$ as a sum of rational functions in which the denominator is a power of a linear polynomial and the numerator is a constant; or the denominator is a power of an irreducible quadratic polynomial and the numerator is linear polynomial.

*To compute $\int\frac{A}{(at+b)^n}\,dt$, $A$ constant, $a\neq 0$, $n$ a positive integer, use the substitution $u=at+b$.

*To compute $\int\frac{At}{(at^2+bt+c)^n}\,dt$ where $at^2+bt+c$ is irreducible quadratic, use the substitution $u=at^2+bt+c$, adjusting the numerator by a constant.

*To compute $\int\frac{A}{at^2+bt+c}\,dt$, with $at^2+bt+c$ irreducible quadratic, complete the square, use a substitution, and use the arctangent.

*To compute $\int\frac{A}{(at^2+bt+c)^n}\,dt$ with $n\gt 1$, $at^2+bt+c$ irreducible quadratic, complete the square, use a change of variable, and use the reduction formula
$$
\int\frac{dx}{(c^2\pm x^2)^n} = \frac{1}{2c^2(n-1)}\left(\frac{x}{(c^2\pm x^2)^{n-1}} + (2n-3)\int\frac{dx}{(c^2\pm x^2)^{n-1}}\right).$$
