Integrate $\int \frac{1}{(1-x)(1+x)}dx$

Question

Integrate $$\int \frac{1}{(1-x)(1+x)}dx$$

$$\int \frac{1}{1-x^2}dx$$

$$=\tanh^{-1}(x)+C$$

When I look on Desmos though, this is only part of the answer?

The blue is the function that it is supposed to be, and the red is the derivative of the answer I got. As you can see it's right, but only the red is shaded, the other two blue regions are not. Why is this? How can I fix this? My answer is correct, right?

Answer 1

$x=\pm 1$ are simple poles for the integrand function, in particular non-integrable singularities. That implies $\int_{a}^{b}\frac{dx}{1-x^2}$ has no meaning if $1$ or $-1$ belong to $[a,b]$. On the other hand, $$ \frac{1}{1-x^2} = \frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)$$ clearly holds for any $x\neq \pm 1$, hence if $1$ and $-1$ do not belong to $[a,b]$ we have $$ \int_{a}^{b}\frac{dx}{1-x^2} = \frac{1}{2}\left(\log|x+1|-\log|x-1|\right).$$ That explains why the depicted primitive only exists in $(-1,1)$.

Answer 2

It looks like you're integrating your function $1/(1-x^2)$ on $[1,2]$. Yet, it's singular at $1$, so any definite integral that goes over $\pm 1$ will not be correct.