求不定积分（x^2+5x+1）^(-1/2)*x^(-1)用换元法和分部积分法都无法解决,

Question

求不定积分（x^2+5x+1）^(-1/2)*x^(-1)
用换元法和分部积分法都无法解决,

Answer 1

令t=1/x 则dx=-dt/t^2
∴∫dx(x^2+5x+1）^(-1/2)*x^(-1)
=-∫dt|t|/[t(t^2+5t+1)^(1/2)]
=-sgn(t)∫dt(t^2+5t+1)^(-1/2)
=-sgn(t)∫dt[(t+5/2)^2-21/4]^(-1/2)
=-sgn(t)In|t+5/2+(t^2+5t+1)^(1/2)|+C
=-sgn(x)In|1/x+5/2+[(x^2+5x+1)^(1/2)]/|x||+C
=-In|1/x+5/2+[(x^2+5x+1)^(1/2)]/x|+C
或令x+5/2=21^(1/2)/2*sect 0