How do you integrate #int 2/((x^4-1)dx# using partial fractions?

Question

Christopher Allers · Accepted Answer

#int 2/(x^4-1) dx=-tan^-1 x-1/2ln(x+1)+1/2ln(x-1)+C_0#

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

for the partial fraction

#2/(x^4-1) =2/((x^2+1)(x^2-1))=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)#

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

We now have the equation

#2=Ax^3+Bx^2-Ax-B+C(x^3+x-x^2-1)+D(x^3+x+x^2+1)#

and then

#A+C+D=0" "#first equation #B-C+D=0" "#second equation #-A+C+D=0" "#third equation #-B-C+D=2" "#fourth equation

simultaneous solution results to

#A=0# #B=-1# #C=-1/2# #D=1/2#

#int 2/(x^4-1) dx =int(Ax+B)/(x^2+1) dx+int C/(x+1) dx+int D/(x-1) dx#

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

#int 2/(x^4-1) dx=-tan^-1 x-1/2ln(x+1)+1/2ln(x-1)+C_0#

God bless....I hope the explanation is useful.

Cameron Alexander · Answer

undefined To integrate the function 2/(x^4 - 1) dx using partial fractions, you first factor the denominator as a difference of squares: (x^2 + 1)(x^2 - 1). Then, you decompose 2/(x^4 - 1) into partial fractions:

2/(x^4 - 1) = A/(x^2 + 1) + B/(x - 1) + C/(x + 1)

Next, you find the values of A, B, and C by multiplying both sides by (x^4 - 1) and simplifying:

2 = A(x^2 - 1)(x + 1) + B(x^2 + 1)(x + 1) + C(x^2 + 1)(x - 1)

After expanding and simplifying, you can equate coefficients of like terms to find A, B, and C.

Finally, once you have found the values of A, B, and C, you integrate each term separately and then sum the results to find the integral of 2/(x^4 - 1) dx.