How to prove that: $\int_{-1}^{1} \frac{1}{x^2-1}dx$ diverges? I have this integral
$$\int_{-1}^{1} \frac{1}{x^2-1}dx$$
I know it diverges but I don't know how to prove it. We have not learned hyperbolic trig functions so I know this is related to inverse hyperbolic tangent but I am looking for another approach that does not rely on that knowledge.
 A: Partial fractions for sure (and logs will emerge), and consider the two improper integrals
$$
\int_{-1}^0 \frac{1}{x^2-1} \ dx \ \text{ with } \int_0^1 \frac{1}{x^2-1} \ dx.
$$
You need BOTH to converge for your integral to converge.  Carry on...
A: Note that for $x\in (-1,1)$,
$$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}\leq \frac{1}{(-2)(x+1)}.$$
Hence
$$\int_{-1}^{1} \frac{1}{x^2-1}dx\leq -\frac{1}{2}\int_{-1}^{1} \frac{1}{x+1}dx=-\left[\frac{\ln(x+1)}{2}\right]_{-1}^1=-\infty.$$
A: Here's a less orthodox answer. Consider just the interval $[0, 1)$, and divide it into intervals $[0, 1/2), [1/2, 2/3), \ldots, [1 - \frac{1}{n}, 1 - \frac{1}{n+1}), \ldots$, each of width $\frac{1}{n(n+1)}$. The value of $\frac{1}{1-x^2}$ at $x = 1 - \frac{1}{n}$ is $\frac{n^2}{2n-1}$. Multiplying widths by function values gives a lower Riemann sum $$\int_0^1 \frac{dx}{1-x^2} > \sum_{n=1}^\infty \frac{n}{(n+1) (2n-1)}$$
which diverges as the harmonic series.
A: First of all, since the integral is improper both at $x=-1$ and at $x=1$, you should split it:
$$
\int_{-1}^1 \frac{dx}{x^2-1} = \int_{-1}^0 \frac{dx}{x^2-1} + \int_0^1 \frac{dx}{x^2-1} = \lim_{a \to -1} \int_a^0 \frac{dx}{x^2-1} + \lim_{b \to 1} \int_0^b \frac{dx}{x^2-1}.
$$
Consider the last integral, which I write as
$$
- \int_0^b \frac{dx}{1-x^2} = -\int_0^b \frac{dx}{(1-x)(1+x)}.
$$
Observe that, for $0<x<b$, $1+x<1+b<1+1=2$. Hence
$$
\int_{0}^b \frac{dx}{(1+x)(1-x)} > \frac{1}{2} \int_0^b \frac{dx}{1-x} = \frac{1}{2} \left[ -\log |1-x| \right]_0^b
$$
Now, let $b \to 1$ and deduce that
$$
\lim_{b \to 1} \int_0^b \frac{dx}{(1+x)(1-x)} = +\infty.
$$
Hence the improper integral diverges.
A: Hin: On the interval $[0,1),$ 
$$\frac{1}{x^2-1}\le \frac{1}{2}\frac{1}{x-1}.$$
If the integral of the right side over $[0,1)$ is $-\infty,$ then the same is true of the left side.
A: [The following is essentially @Guy Fsone's answer. It is rewritten differently.]
Observe that
$$
\frac{1}{x^2-1} = -\frac{1}{2}\left(\frac{1}{x+1} +\frac{1}{1-x}\right)= -\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)'
$$
Hence, 
$$
\int_{-1}^{1} \frac{1}{x^2-1}dx=\left[-\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right]^1_{-1}=\frac{1}{2}[\ln(1-x)-\ln(1+x)]\big|_{-1}^1
\tag{1}
$$
But note that, 
$$
\lim_{x\to 0^+} \ln\left(x\right)=-\infty \tag{2}
$$
Combining (1) and (2) one has
$$
\int_{-1}^{1} \frac{1}{x^2-1}\ dx=\frac{1}{2}\left[(-\infty-\ln 2)-\big(\ln 2-(-\infty)\big)\right]=-\infty
$$
where we use the arithmetic operations of the extended real number line. 
