#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.
