How do you find #int (x^3-x-1)/((x^4-1)x) dx# using partial fractions?

Question

Aurora Alvarado · Accepted Answer

#Lnx-1/4*Ln(x^4-1)+arctanx+C#

I decomposed integrand into basic fractions,

#(x^3-x-1)/[x*(x^4-1)]#

=#(x^3-x-1)/[x(x+1)(x-1)(x^2+1)]#

=#A/x+B/(x+1)+C/(x-1)+(Dx+E)/(x^2+1)#

After expanding denominators,

#A(x+1)(x-1)(x^2+1)+Bx(x-1)(x^2+1)+Cx(x+1)(x^2+1)+(Dx+E)x(x^2-1)=x^3-x-1#

Set #x=-1#, #4B=-1 or B=-1/4#

Set #x=0#, #-A=-1 or A=1#

Set #x=1#, #4C=-1 or C=-1/4#

Set #x=i#, #(E+Di)*(-2i)=-1-2i or E+Di=1-1/2*i#. So #D=-1/2 and E=1#

Hence,

#(x^3-x-1)/[x*(x^4-1)]#

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

Hence,

#int (x^3-x-1)/[x*(x^4-1)]*dx#

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

=#Lnx-1/4*Ln(x+1)-1/4*Ln(x-1)#-#1/4#*#int (2x*dx)/(x^2+1)#+#dx/(x^2+1)#

=#Lnx-1/4*Ln(x+1)-1/4*Ln(x-1)-1/4*Ln(x^2+1)+arctanx+C#

=#Lnx-1/4*Ln(x^4-1)+arctanx+C#

Adam Adkins · Answer

undefined To find $ \int \frac{x^3 - x - 1}{(x^4 - 1)x} dx $ using partial fractions, we first factor the denominator $ x^4 - 1 $ as $ (x^2 + 1)(x + 1)(x - 1) $. Then, we express the given fraction as a sum of simpler fractions with denominators of $ x^2 + 1 $, $ x + 1 $, and $ x - 1 $. The general form of the partial fraction decomposition is:

$$ \frac{x^3 - x - 1}{(x^4 - 1)x} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1} + \frac{Ex + F}{x - 1} + \frac{Gx + H}{x + 1} $$

Multiplying both sides by $ (x^4 - 1)x $ to clear the fractions, we get:

$$ x^3 - x - 1 = A(x^3 - x) + B(x^2 - 1) + (Cx + D)(x^3 - x) + (Ex + F)(x^2 + 1) + (Gx + H)(x^2 + 1) $$

Now, we can equate coefficients of like terms on both sides to solve for $ A $, $ B $, $ C $, $ D $, $ E $, $ F $, $ G $, and $ H $. Once we find these values, we can integrate each term individually to find the integral of the original function.