Question

How can I integrate `int 1/(x^4+x) dx` using partial fractions?

My problem is that I don't know what will be in the numerator over the `x^3+1`
`A` or `Ax+B`? I've learnt that only these two options exist. But I also saw `Ax^2+B` so I'm kinda confused.

Answer 1

`int (dx)/(x^4+x)=Lnx-1/3ln(x^3+1)+C`

Explanation:

Now I decomposed integrand into basic fractions,

`int (dx)/(x^4+x)`

=`int (dx)/[(x^3+1)*x]`

=`int ((x^3+1)*dx)/[(x^3+1)*x]`-`int (x^3*dx)/[(x^3+1)*x]`

=`int (dx)/x`-`int (x^2*dx)/(x^3+1)`

=`int (dx)/x`-`1/3*int (3x^2*dx)/(x^3+1)`

=`Lnx-1/3Ln(x^3+1)+C`

Answer 2

The answer is `=ln(|x|)-1/3ln(|x+1|)-1/3ln(|x^2-x+1|)+C`

Explanation:

Reminder

`a^3+b^3=(a+b)(a^2-ab+b^2)`

The denominator is

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

Therefore, the decomposition into partial fractions is

`1/(x^4+x)=A/(x)+B/(x+1)+(Cx+D)/(x^2-x+1)`

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

The denominators are the same, compare the numerators

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

Compare the LHS and the RHS

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

Let `x=-1`, `=>`, `1=B(-1)()`, `=>`, `B=-1/3`

Coefficients of `x^3`

`0=A+B+C`

`C=-A-B=-1+1/3=-2/3`

Coefficients of `x^2`

`0=-B+C+D`

`D=B-C=-1/3+2/3=1/3`

Therefore,

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

So,

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

`=ln(|x|)-1/3ln(|x+1|)-1/3ln(|x^2-x+1|)+C`