Question

求积分∫[2x+6/x^3+x-2]dx

Answer 1

观察出x³+x-2有一个因式为x-1,

x³+x-2=x³-x²+x²-x+2x-2=x²(x-1)+x(x-1)+2=(x-1)(x²+x+2)

则：(2x+6)/(x³+x-2)

=(2x+6)/[(x-1)(x²+x+2)]

=A/(x-1)+(Bx+C)/(x²+x+2)

合并后比较系数得：A=2,B=C=-2

则：(2x+6)/(x³+x-2)=2/(x-1)-(2x+2)/(x²+x+2)

∫[(2x+6)/(x³+x-2)]dx

=2∫1/(x-1)dx - ∫ (2x+2)/(x²+x+2)dx

=2ln|x-1| - ∫ (2x+1)/(x²+x+2)dx - ∫ 1/(x²+x+2)dx

=2ln|x-1| - ∫ 1/(x²+x+2)d(x²+x) - ∫ 1/[(x+1/2)²+7/4]dx

=2ln|x-1| - ln(x²+x+2) - (2/√7)arctan[(x+1/2)/√7] + C