定积分计算

更新时间2019-01-24 14:36:07

【1】1/(x^4-x²+1)=1/[(x²+2x²+1)-3x²]

=1/[(x²+1)²-(x√3)²]=1/[(x²-x√3+1)(x²+x√3+1)]

=[(x+√3)/(x²+x√3+1)-(x-√3)/(x²-x√3+1)]/(2√3)

【21】f(x)=∫(x+√3)dx/[(2√3)(x²+x√3+1)]

=∫d(x²+x√3)/[(4√3)(x²+x√3+1)]+∫dx/[(2x+√3)²+1]

=[1/(4√3)]ln(x²+x√3+1)+(1/2)arctan(2x+√3)+c'

【22】同理g(x)=∫(x-√3)dx/[(2√3)(x²-x√3+1)]

=[1/(4√3)]ln(x²-x√3+1)-(1/2)arctan(2x-√3)+c"

【2】F(x)=∫dx/(x^4-x²+1)=f(x)-g(x)

=[1/(4√3)]ln[(x²+x√3+1)/(x²-x√3+1)]

+(1/2)[arctan(2x+√3)+arctan(2x-√3)]+c

【3】I=F(∞)-F(-∞)

=lim[arctan(2x+√3)+arctan(2x-√3)]

  x→+∞

=π/2+π/2

=π