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#