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Mathematics Question on integral

Integrate the rational function: 2(1−x)(1+x2)\frac {2}{(1-x)(1+x^2)}

Answer

Let 2(1−x)(1+x2)\frac {2}{(1-x)(1+x^2)} == A(1−x)+Bx+C(1+x2)\frac {A}{(1-x)}+\frac {Bx+C}{(1+x^2)}

2=A(1+x2)+(Bx+C)(1−x)2 = A(1+x^2)+(Bx+C)(1-x)

2=A+Ax2+Bx−Bx+C−Cx2 = A+Ax^2+Bx-Bx+C-Cx

Equating the coefficient of x2, x, and constant term, we obtain
A−B=0A − B = 0
B−C=0B − C = 0
A+C=2A + C = 2
On solving these equations, we obtain
A=1, B=1, and C=1A = 1, \ B = 1, \ and \ C = 1

∴ 2(1−x)(1+x2)\frac {2}{(1-x)(1+x^2)} = 11−x\frac {1}{1-x} + x+11+x2\frac {x+1}{1+x^2}

⇒ ∫$$\frac {2}{(1-x)(1+x^2)} = ∫$$\frac {1}{1-x}\ dx+ ∫$$\frac {x}{1+x^2}\ dx + ∫$$\frac {1}{1+x^2}\ dx

                                  = - $∫$$\frac {1}{1-x}\ dx$ + $\frac 12$$∫$$\frac {2x}{1+x^2}\ dx$ + $∫$$\frac {1}{1+x^2}\ dx$

                                  = -$log\ |x-1|+\frac 12log|1+x^2|+tan^{-1}x+C$