Evaluate: ∫2(1−x)(1+x²)dx

Let I = ∫2(1−x)(1+x²)dx
2(1−x)(1+x²) = A/(1−x) + (Bx+C)/(1+x²)
⇒ 2(1−x)(1+x²) = A(1+x²) + (Bx+C)(1−x)
(1−x)(1+x²) ⇒ 2 = A(1+x²) + (Bx+C)(1−x)
⇒ 2 = A + Ax² + Bx − Bx² + C − Cx
⇒ 2 = (A+C) + (A−B)x² + (B−C)x
Equating coefficients both sides, we get
A+C = 2, A−B = 0, B−C = 0
⇒ A = B, B = C ⇒ A = B = C, A+A = 2 ⇒ A = 1
therefore A = B = C = 1
So, 2(1−x)(1+x²) = 1/(1−x) + (x+1)/(1+x²)
∫2(1−x)(1+x²) = ∫1/(1−x)dx + ∫(x+1)/(1+x²)dx = ∫1/(1−x)dx + ∫x/(1+x²)dx + ∫1/(1+x²)dx
= −log|1−x| + (1/2)log|1+x²| + tan⁻¹x + C