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.
 A: 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
$$
\int\frac{dx}{(c^2\pm x^2)^n} = \frac{1}{2c^2(n-1)}\left(\frac{x}{(c^2\pm x^2)^{n-1}} + (2n-3)\int\frac{dx}{(c^2\pm x^2)^{n-1}}\right).$$
A: Use $u = t+1$. Then $t = u-1$ and $du = dt$, so that 
$$
\int \frac t{t+1} dt = \int \frac{u-1}u du = \int 1 du - \int \frac 1u du = u - \log u + C = t+1 - \log(t+1) + C.
$$
"Unintegrable functions" are not functions with no primitive, they're just functions that we don't know how to compute the primitive by hand, i.e. they're not "expressible in terms of elementary functions".
Hope that helps,
A: $u= t+1$ : $du=dt$ $\rightarrow \displaystyle \int\frac{u-1}{u} du = \int\frac{u}{u}\;du- \int\frac{1}{u}\;du = u-\ln u = t+1 - \ln(t+1) + C$ 
Hope I helped. 
