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Mathematics Question on integral

Integrate the rational function: 1x4−1\frac {1}{x^4-1}

Answer

1x4−1\frac {1}{x^4-1} = 1(x2−1)(x2+1)\frac {1}{(x^2-1)(x^2+1)}= 1(x+1)(x−1)(1+x2)\frac {1}{(x+1)(x-1)(1+x^2)}

Let 1(x+1)(x−1)(1+x2)\frac {1}{(x+1)(x-1)(1+x^2)} = A(x+1)+B(x−1)+Cx+D(x2+1)\frac {A}{(x+1)}+\frac {B}{(x-1)}+\frac {Cx+D}{(x^2+1)}

1=A(x−1)(x2+1)+B(x+1)(x2+1)+(Cx+D)(x2−1)1 = A(x-1)(x^2+1)+B(x+1)(x^2+1)+(Cx+D)(x^2-1)

1=A(x3+x−x2−1)+B(x3+x+x2+1)+Cx3+Dx2−Cx−D1 = A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+Cx^3+Dx^2-Cx-D

1=(A+B+C)x3+(−A+B+D)x2+(A+B−C)x+(−A+B−D)1 = (A+B+C)x^3+(-A+B+D)x^2+(A+B-C)x+(-A+B-D)

Equating the coefficient of x3,x2,x,and constant term, we obtainEquating \ the\ coefficient\ of\ x^3 , x^2 , x, and \ constant \ term, \ we \ obtain

A+B+C=0A+B+C = 0
−A+B+D=0-A+B+D = 0
A+B−C=0A+B-C = 0
−A+B−D=1-A+B-D = 1
On solving these equations, we obtainOn\ solving\ these\ equations, \ we \ obtain
A=−14, B=14, C=0,and D=−12A=-\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 = −14 log∣x−1∣+14log ∣x−1∣−12 tan−1x+C-\frac 14\ log|x-1|+\frac 14log\ |x-1|-\frac 12\ tan^{-1}x+C

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