∫（1／x（2x²+1））dx

∫（1／x（2x²+1））dx
∫（1／x（2x²+1））dx

∫（1／x（2x²+1））dx
∫ 1/[x(2x² + 1)] dx
= ∫ 1/x dx - 2∫ x/(2x² + 1) dx
= ln|x| - 2∫ 1/(2x² + 1) d(x²/2)
= ln|x| - (1/2)∫ 1/(2x² + 1) d(2x²)
= ln|x| - (1/2)∫ 1/(2x² + 1) d(2x² + 1)
= ln|x| - (1/2)ln(2x² + 1) + C
拆解步骤：
1/[x(2x² + 1)] = A/x + (Bx + C)/(2x² + 1)
1 = A(2x² + 1) + (Bx + C)x = (2A + B)x² + Cx + A
{ A = 1,C = 0
{ 2A + B = 0 => B = -2A = -2
所以1/[x(2x² + 1)] = 1/x - 2x/(2x² + 1)