Question

How do you integrate `(x^2+x-1)/(x(x^2-1))` using partial fractions?

Answer 1

The answer is `=ln(∣x∣)-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C`

Explanation:

We use

`a^2-b^2=(a+b)(a-b)`

So, the denominator is

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

The decomposition in partial fractions is

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

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

So,

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

Let, `x=0`,`=>`, `-1=-A`, `=>`, `A=1`

Let `x=1`, `=>`, `1=2C`, `=>`, `C=1/2`

Let `x=-1`,`=>`, `-1=2B`, `=>`, `B=-1/2`

So,

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

So,

`int((x^2+x-1)dx)/(x(x+1)(x-1))=intdx/x+int(-1/2dx)/(x+1)+int(1/2dx)/(x-1)`

`=ln(∣x∣)-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C`

Answer 2

Narad is correct. Alternatively use the cover-up rule to get `A`, `B` and `C` directly.

Explanation:

To get `A` "cover up" the `x` in the denominator and replace all other `x`'s with value of `x` which makes the covered-up term zero, which in this case `x=0`, so `A=(0^2+0-1)/((0+1)(0-1))=1`
and similarly `B` and `C` by putting `x=1` and `x=-1`.