Question

Integrate the rational function: \(\frac {1}{x^4-1}\)

Answer 1

\(\frac {1}{x^4-1}\) = \(\frac {1}{(x^2-1)(x^2+1)}\)= \(\frac {1}{(x+1)(x-1)(1+x^2)}\)

Let \(\frac {1}{(x+1)(x-1)(1+x^2)}\) = \(\frac {A}{(x+1)}+\frac {B}{(x-1)}+\frac {Cx+D}{(x^2+1)}\)

\(1 = A(x-1)(x^2+1)+B(x+1)(x^2+1)+(Cx+D)(x^2-1)\)

\(1 = A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+Cx^3+Dx^2-Cx-D\)

\(1 = (A+B+C)x^3+(-A+B+D)x^2+(A+B-C)x+(-A+B-D)\)

\(Equating \ the\  coefficient\  of\  x^3 , x^2 , x, and \ constant \ term, \ we \ obtain\)

\(A+B+C = 0\)
\(-A+B+D = 0\)
\(A+B-C = 0\)
\(-A+B-D = 1\)
\(On\  solving\ these\  equations, \ we \ obtain\)
\(A=-\frac 14, \ B=\frac 14,\ C=0,and \ D=-\frac 12\)

∴ \(\frac {1}{x^4-1}\)\(=-\frac 14(x+1)+\frac 14(x-1)-\frac 12(x^2+1)\)

⇒ \(∫\)\(\frac {1}{x^4-1}\ dx\) = \(-\frac 14\ log|x-1|+\frac 14log\ |x-1|-\frac 12\ tan^{-1}x+C\)

\(=\frac 14\ log|\frac {x-1}{x+1}|-\frac 12tan^{-1} x+C\)