1/【(1+x)(1+x∧2)】的不定积分
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∫ dx/[(1 + x)(1 + x²)]
= (1/2)∫ dx/(1 + x) + (1/2)∫ (1 - x)/(1 + x²) dx
= (1/2)∫ dx/(1 + x) + (1/2)∫ dx/(1 + x²) - (1/2)∫ x/(1 + x²) dx
= (1/2)∫ d(1 + x)/(1 + x) + (1/2)∫ dx/(1 + x²) - (1/4)∫ d(1 + x²)/(1 + x²)
= (1/2)ln(1 + x) + (1/2)arctanx - (1/4)ln(1 + x²) + C
= (1/2)arctanx + ln[√(1 + x)/(1 + x²)^¼] + C
分式分裂过程看下面：
_________________________________________________
令1/[(1 + x)(1 + x²)] = A/(1 + x) + (Bx + C)/(1 + x²)
1 = A(1 + x²) + (Bx + C)(1 + x)
1 = A + Ax² + Bx + Bx² + Cx + C
1 = (A + B)x² + (B + C)x + (A + C)
{ A + B = 0 → B = - A
{ B + C = 0 → C = - B = A
{ A + C = 1 → C = 1 - A
C = 1 - A → A = 1 - A → 2A = 1 → A = 1/2 = C,B = - 1/2
= (1/2)∫ dx/(1 + x) + (1/2)∫ (1 - x)/(1 + x²) dx
= (1/2)∫ dx/(1 + x) + (1/2)∫ dx/(1 + x²) - (1/2)∫ x/(1 + x²) dx
= (1/2)∫ d(1 + x)/(1 + x) + (1/2)∫ dx/(1 + x²) - (1/4)∫ d(1 + x²)/(1 + x²)
= (1/2)ln(1 + x) + (1/2)arctanx - (1/4)ln(1 + x²) + C
= (1/2)arctanx + ln[√(1 + x)/(1 + x²)^¼] + C
分式分裂过程看下面：
_________________________________________________
令1/[(1 + x)(1 + x²)] = A/(1 + x) + (Bx + C)/(1 + x²)
1 = A(1 + x²) + (Bx + C)(1 + x)
1 = A + Ax² + Bx + Bx² + Cx + C
1 = (A + B)x² + (B + C)x + (A + C)
{ A + B = 0 → B = - A
{ B + C = 0 → C = - B = A
{ A + C = 1 → C = 1 - A
C = 1 - A → A = 1 - A → 2A = 1 → A = 1/2 = C,B = - 1/2
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