Question

Express 4x^2-x+3 / (x+1)(x^2-1) as partial fractions?

Answer 1

`(4x^2-x+3)/[(x+1)*(x^2-1)]=3/2*1/(x-1)+5/2*1/(x+1)-4/(x+1)^2`

Explanation:

`(4x^2-x+3)/[(x+1)*(x^2-1)]`

=`(4x^2-x+3)/[(x+1)^2*(x-1)]`

=`A/(x-1)+B/(x+1)+C/(x+1)^2`

After expanding denominator,

`A*(x+1)^2+B*(x^2-1)+C*(x-1)=4x^2-x+3`

Set `x=-1`, hence `-2C=8`, so `C=-4`

Set `x=1`, hence `4A=6`, so `A=3/2`

Set `x=0`, hence `A-B-C=3`, so `B=5/2`

Thus,

`(4x^2-x+3)/[(x+1)*(x^2-1)]=3/2*1/(x-1)+5/2*1/(x+1)-4/(x+1)^2`

Answer 2

`color(blue)(5/(2x+2)-4/((x+1)^2)+3/(2x-2))`

Explanation:

`(4x^2-x+3)/((x+1)(x^2-1))`

Rewrite as:

`(4x^2-x+3)/((x+1)^2(x-1))`

This was formed by exploiting the difference of two squares:

`(x^2-1)=(x+1)(x-1)`

For this we expect the partial form to be:

`(4x^2-x+3)/((x+1)(x^2-1))-=A/(x+1)+B/(x+1)^2+C/(x-1)`

Add RHS:

`(4x^2-x+3)/((x+1)(x^2-1))-=(A(x+1)(x-1)+B(x-1)+C(x+1)^2)/((x+1)^2(x-1))`

Since this is an identity, it follow that the coefficients will be identical. We use this idea to find the values of A, B and C. by equating coefficients:

Coefficients of `x^2`

`[4]=A+C`

Coefficients of `x`

`[-1]=B+2C`

Constant:

`[3] = C-A-B`

We now have to solve:

`A+C=4` `\ \ \ \ \ \ \ \ \ \ [1]`

`B+2C=-1` `\ \ \ \ [2]`

`C-A-B=3` `\ \ \ \[3]`

Finding `[1]` and `[2]` in terms of C

`A=4-C`

`B=-1-2C`

Substituting in `[3]`

`C-(4-C)-(-1-2C)=3`

`C=3/2`

Substituting in `[1]`

`A+3/2=4`

`A=5/2`

Substituting in `[2]`

`B+2(3/2)=-1`

`B=-4`

So we have our values:

`A=5/2, B=-4 , C=3/2`

Substituting these in:

`A/(x+1)+B/(x+1)^2+C/(x-1)`

`5/(2(x+1))-4/((x+1)^2)+3/(2(x-1)`

`color(blue)(5/(2x+2)-4/((x+1)^2)+3/(2x-2))`

It should be remembered that the partial fraction form is not unique to a given fraction. There are different ways it can be expressed as partial fractions.

The key to solving these is knowing which form the partial fraction will be in. This need to be memorised.

You can find the forms they take online or in a suitable text book.