Evaluate $\int {2x\over x^2-1}dx$ My friend evaluated this before he went to bed: $$\int {2x\over x^2-1}dx$$
The answer was $\log(x^2-1)$.
I just can't figure out how that works. I know that $\int \frac1x dx = \log|x|$, so what just happened to $2x$?
 A: $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac1{x-1}+\frac1{x+1}$

Alternatively, using Partial Fraction Decomposition, let $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac A{x+1}+\frac B{x-1}$ where $A,B$ are arbitrary constants.
So, $2x=(A+B)x+B-A$
Comparing the constant terms in either of the identity, $B-A=0\implies A=B$ 
Comparing the coefficients of $x,A+B=2\implies A=B=1$
So, $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac 1{x+1}+\frac 1{x-1}$
A: $\dfrac{2x}{x^2-1}dx=\dfrac{d(x^2-1)}{x^2-1}$
A: Substitute $u=x^2-1$, $\mathrm du=2x\,\mathrm dx$
Then $\int \frac{2x}{x^2-1}\,\mathrm dx=\int \frac{1}{u} \,\mathrm du=\log(u)=\log(x^2-1)$
A: $\int {2x\over x^2-1}dx
= -\int {2x\over 1-x^2}dx
= -\int 2x dx \sum_{n=0}^{\infty} x^{2n}
= -2\int dx \sum_{n=0}^{\infty} x^{2n+1}
= -2\sum_{n=0}^{\infty}\int x^{2n+1}dx
= -2\sum_{n=0}^{\infty} {x^{2n+2} \over 2n+2}
= -\sum_{n=0}^{\infty} {(x^2)^{n+1} \over n+1}
= -\sum_{n=1}^{\infty} {(x^2)^{n} \over n}
= \ln(1-x^2)
$.
Check:
$(\ln(1-x^2))'
= {-2x \over 1-x^2}
= {2x \over x^2-1}
$.
Just playing with power series Eulerishly to see what would happen.
A: Let $\quad\quad u  = \;x^2 - 1.\quad$ (Use "$\;u$-substitution.")
Then $\;\;\; du = 2x \;\;dx$. 
$($Recall, we need to replace $\;2x\;dx\;$ with $\;du\;$ to integrate in terms of $u.)$ 
This gives us...
$$\int \frac{2x \;dx}{x^2 - 1} \quad=\quad \int \frac{du}{u} \quad= \quad\int \frac{1}{u} du \;\;= \;\;\;?$$
From what you stated in your question, I think you can go from here?
A: $$\int\frac{2x}{x^2-1}dx=\int\frac{(x+1)+(x-1)}{x^2-1^2}dx=$$
$$=\int\frac{(x+1)+(x-1)}{(x+1)(x-1)}dx=\int\left(\frac{x+1}{(x+1)(x-1)}+\frac{x-1}{(x+1)(x-1)}\right)dx=$$
$$=\int\left(\frac1{x-1}+\frac1{x+1}\right)dx=\int\left(\frac1{x-1}\right)dx+\int\left(\frac1{x+1}\right)dx=$$
$$=\ln(x-1)+\ln(x+1)+C=\ln(x-1)(x+1)+C=\ln(x^2-1)+C$$
A: In general, $$\int(\frac{\frac{d}{dx}(f(x))}{f(x)}dx=\log(|f(x)|)+C$$
(As seen in $\frac1x dx = \log|x|+C$)
Similarly, $$\frac{d}{dx}(x^2-1)=2x$$
$$\therefore\int {2x\over x^2-1}dx=\int(\frac{\frac{d}{dx}(x^2-1)}{x^2-1}dx$$
$$=\log(|x^2-1|)+C$$
We apply mod in logarithm to take only the positive value of $x^2-1$ because if $x<1$ then $(x^2-1)<0$ and logarithms of negative values do not exist.
