3
$\begingroup$

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?

enter image description here

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?

  • 0
    You can integrate this simply by doing a partial fraction decomposition on $\frac{1}{(1-x)(1+x)}$. Also, your integral is missing a $dx$! – 2017-02-07
  • 1
    I know I can, but why is this wrong – 2017-02-07
  • 0
    As a general advice: dont trust too much on desmos on any other graphical or CAS software. Desmos fail sometimes representing functions. – 2017-02-07

2 Answers 2

5

$x=\pm 1$ 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 $(-1,1)$.

0

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.