Why $\ln (x)-\ln(x)$ as $x$ approaches infinity is not zero? The questions itself is in "improper integrals" subject.
$$\int \frac{2x}{x^{2}-1}$$
Here is an antiderivative: $\ln(x^{2}-1)|_{x=-1}^{x=1}$
If I just place numbers instead of $x$ I get $\ln(0)-\ln(0)$ which is like writing number-number=zero. The teacher however said that this was a mistake and that it is not based on same principle.
why? is it not like substracting a number from itself?
 A: It is true that $\ln(x^2-1)$ is an antiderivative of $\dfrac{2x}{x^2-1}$ on an interval where $\ln(x^2-1)$ is defined and differentiable.  But unless you want to get into complex values for logarithms of negative numbers, you won't want to use this for $-1 < x < 1$.  Instead, for $-1 < x < 1$ you could take the antiderivative as $\ln(1-x^2)$.
Then $$\int_a^b \dfrac{2x}{x^2-1} \; dx = \ln(1-b^2) - \ln(1-a^2)$$
for $-1 < a < b < 1$.  But you can't use this for $a=-1$ and $b=1$, because
$\ln(0)$ is undefined.  In fact, you can't take the limit as $a \to -1$ and $b \to 1$, because $\ln(z) \to \infty$ as $z \to 0+$.  Now you might say, why not take
$$ \lim_{b \to 1-}\int_{-b}^b \dfrac{2x}{1-x^2}\; dx  = \lim_{b \to 0+} 0 = 0$$
The trouble is that if you do the limit differently, with $a \ne -b$, you'll get a different answer.  For example, 
$$ \lim_{b \to 1-} \int_{-b^2}^b \dfrac{2x}{1-x^2}\; dx = \lim_{b \to 1-} \ln(1+b^2) = \ln 2 $$
So all we can say is 
$$ \int_{-1}^1 \dfrac{2x}{x^2-1}\; dx \ \text{diverges} $$
A: The map $x\mapsto\ln(x^2-1)$ is an antiderivative of $\dfrac{2x}{x^2-1}$. But if you replace $x$ by to concrete numbers and compute their difference, all you get as number, not an antiderivative of that function. Besides, $\ln(0)$ is undefined.
However, it is true that$$\lim_{t\to1^-}\int_{-t}^t\frac{2x}{x^2-1}\,\mathrm dx=0.$$That is so because, for each $t\in(0,1)$,$$\int_{-t}^t\frac{2x}{x^2-1}\,\mathrm dx=\ln(t^2-1)-\ln\bigl((-t)^2-1\bigr)=0.$$
