Partial Fractions

**dadon** · April 15th 2006, 05:55 AM

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

**TD!** · April 15th 2006, 07:05 AM

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.

**dadon** · April 16th 2006, 01:11 AM

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

**CaptainBlack** · April 16th 2006, 01:49 AM

Originally Posted by dadon

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

**dadon** · April 16th 2006, 03:13 AM

Thank you!

You guys are certainly good at maths.

Regards

dadon

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