How do you integrate #int x^4/(x^4-1)# using partial fractions?

Since the denominator is the same as the numerator, you need to do long division first:

#(x^4-1)sqrt(x^4)#

=#1+1/(x^4-1)#

Then, we started to do partial fraction.
Since the denominator is not a linear function, we need to check it if it is reducible. Therefore, the denominator is reduced to:

#x^4-1=(x^2+1)(x^2-1)#

These factors are also not linear function. Therefore, we check again whether these funtion can be reduce and we find out:

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

So, overall the denominator is:

#x^4-1=(x^2+1)(x+1)(x-1)#

Then, we do partial fraction:

#1/((x^2+1)(x+1)(x-1))=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)#

#1=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)#

subtitute #x=1# into the equation

#1=4D#
#D=1/4#

subtitute #x=-1# into the equation

#1=-4C #
#C=-1/4#

compare the coefficient of #x^3#on both side

#0=A+C+D#
#A=-C-D#
# =1/4-1/4#
#A=0#

compare the constant number on both side

#1=-B-C+D#
#B=-1-C+D#
#=-1+1/4+1/4#
#B=-1/2#

so

#intx^4/(x^4-1)=int(1-1/(2(x^2+1))-1/(4(x+1))+1/(4(x-1)))#

#intx^4/(x^4-1)=x-int1/2*1/(x^2+1)-int1/4*1/(x+1)+int1/4*1/(x-1)#

#intx^4/(x^4-1)=x-1/2int1/(x^2+1)-1/4int1/(x+1)+1/4int1/(x-1)#

use formula :

#int(f'(x))/(f(x))=Inf(x)#

#intx^4/(x^4-1)=x-1/2arctan(x)-1/4ln(x+1)+1/4ln(x-1)#

Sometimes we can obtain Partial fraction with simple adjustment methods, without using A, B, C..etc.The advantage of this method is : no need to solve for A,B ,C...etc

Here,

#x^4/(x^4-1)=((x^4color(violet)(-1))+color(violet)(1))/(x^4-1)= (x^4-1)/(x^4-1)+1/(x^4-1)#

#=>x^4/(x^4-1)=1+1/(x^4-1)#

#=>x^4/(x^4-1)=1+1/((x^2-1)(x^2+1)#

Now,

#color(blue)((x^2+1)-(x^2-1)=2#

So,

#x^4/(x^4-1)=1+1/2[color(blue)(2)/((x^2-1)(x^2+1))]#

#x^4/(x^4-1)=1+1/2[color(blue)(((x^2+1)-(x^2-1)))/((x^2-1)(x^2+1))]#

#x^4/(x^4-1)=1+1/2[(x^2+1)/((x^2-1)(x^2+1))-(x^2-1)/((x^2-1) (x^2+1))]#

#x^4/(x^4-1)=1+1/2[1/(x^2-1)-1/(x^2+1)]#

Integrating each term we get

#I=intx^4/(x^4-1)dx=int1dx+1/2int[color(red)(1/(x^2-1))-color(brown)(1/(x^2+1))]dx#

#I=x+1/2[color(red)(1/2ln|(x-1)/(x+1)|)-color(brown)(tan^-1x)]+c#

#I=x+1/4ln|(x-1)/(x+1)|-1/2tan^-1x+c#

Note:

#color(red)((1)int1/(x^2-a^2)dx=1/(2a)ln|(x-a)/(x+a)|+c#

#color(brown)((2)int1/(x^2+a^2)dx=1/atan^-1(x/a)+c#