# Partial Fractions

• Apr 15th 2006, 05:55 AM
Partial Fractions
Express $\frac{x^2 + 1}{x^2 - 1}$ in partial fractions and hence evaluate $\int\frac{x^2 + 1}{x^2 - 1} dx$
• Apr 15th 2006, 07:05 AM
TD!
First do the division, or use this trick:

$
\frac{{x^2 + 1}}{{x^2 - 1}} = \frac{{x^2 - 1 + 2}}{{x^2 - 1}} = 1 + \frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}
$

Now do partial fractions on the last one.
• Apr 16th 2006, 01:11 AM
What do you mean by do partial fractions on the last one?
• Apr 16th 2006, 01:49 AM
CaptainBlack
Quote:

What do you mean by do partial fractions on the last one?

He means write:

$
\frac{2}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}
$

Then:

$
\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:

$A+B=0$, and $A-B=2$, which means $A=1$ and $B=-1$, and hence:

$
\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:

$
\int\frac{x^2 + 1}{x^2 - 1} dx=\int 1 + \frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}dx=$
$\int \{1+\frac{1}{x-1}-\frac{1}{x+1}\} dx
$
,

which can be integrates easily.

RonL
• Apr 16th 2006, 03:13 AM