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.

2 Answers
Dec 14, 2017

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

Dec 15, 2017

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