1. ## Partial Fractions

Express $\frac{x^2 + 1}{x^2 - 1}$ in partial fractions and hence evaluate $\int\frac{x^2 + 1}{x^2 - 1} dx$

2. 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.

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

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

5. ## Re:

Thank you!

You guys are certainly good at maths.

Regards