Express $\displaystyle \frac{x^2 + 1}{x^2 - 1}$ in partial fractions and hence evaluate $\displaystyle \int\frac{x^2 + 1}{x^2 - 1} dx$
He means write:Originally Posted by dadon
$\displaystyle
\frac{2}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}
$
Then:
$\displaystyle
\frac{2}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}=\frac{(A+B)x+(A-B)}{(x-1)(x+1)}
$.
Hence:
$\displaystyle A+B=0$, and $\displaystyle A-B=2$, which means $\displaystyle A=1$ and $\displaystyle B=-1$, and hence:
$\displaystyle
\frac{2}{(x-1)(x+1)}=\frac{1}{x-1}-\frac{1}{x+1}$.
As each of the terms in this are easily to integrate your integral may be written:
$\displaystyle
\int\frac{x^2 + 1}{x^2 - 1} dx=\int 1 + \frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}dx=$$\displaystyle \int \{1+\frac{1}{x-1}-\frac{1}{x+1}\} dx
$,
which can be integrates easily.
RonL