guideline for splitting this fraction

We have:

2/(x^2-1)(x^2+1)

by looking at this, what steps do I have to take to find out that it is equal to:

(1/(x^2-1)) - (1/(x^2+1))

?

You see I had to integrate this, and i did it by factoring completely the denominator then using partial fraction decomposition and i did get the right answer, however the book solution did not use PFD, and instead does that 'splitting of the fraction' as shown above, which then makes it rather easy to integrate then and there now that its been split since right away I can tell that it is an inverse tanh and an inverse tan, so it seems to involve less steps which saves time and effort.

Re: guideline for splitting this fraction

The idea is just the same as "partial fractions".

Write

Multiply both sides by to get .

That must be true for all x so choosing two values for x will give you two equations to solve for x.

For example, taking x= 0 gives A- B= 2 and taking x= 1 give 2= 2A.

Or combine "like powers" to write it as . Since that is to be true for all x, you must have A+ B= 0 and A- B= 2.

Re: guideline for splitting this fraction

you can write x^2-1=(x-1)(x+1) and u would have:

2/(x-1)(x+1)(x^2+1)=a/(x-1)+b/(x+1)+(cx+d)/(x^2+1)

cx+d comes because x^2+1 is a quadratic polynomial

find a,b,c,d

Re: guideline for splitting this fraction

Quote:

Originally Posted by

**prasum** you can write x^2-1=(x-1)(x+1) and u would have:

2/(x-1)(x+1)(x^2+1)=a/(x-1)+b/(x+1)+(cx+d)/(x^2+1)

cx+d comes because x^2+1 is a quadratic polynomial

find a,b,c,d

That is exactly what I did.. but in my question I am asking how to decompose the fraction in a different manner which results in

(1/(x^2-1)) - (1/(x^2+1))

so I solve only for A and B equalling to 1 and -1,

rather than going through

a/(x-1)+b/(x+1)+(cx+d)/(x^2+1)

Halls of Ivy thank you for showing me how to decompose it in that way.

My point here was to have less steps in reaching the final integrated simplified form.. the (1/(x^2-1)) - (1/(x^2+1)) form automatically allows one to identify the case for inverse tanh and inverse tan

Maybe I shouldve just posted this in calculus section.. next time i will, sorry