Question

dx/x根号x^2+4x+3详解啊,快

Answer 1

I = ∫dx/[x√(x^2+4x+3)] = ∫dx/[x√(x+2)^2-1)]
令 x+2=sect,则 x=sect-2,
I =∫secttantdt/[(sect-2)tant] =∫sectdt/(sect-2)
=∫dt/(1-2cost)
令 tan(t/2)=u,则 dt=2du/(1+u^2),cost=(1-u^2)/(1+u^2),
I = ∫dt/(1-2cost) = ∫2du/(3u^2-1)
=∫du/(√3u-1)-∫du/(√3u+1)
= (1/√3)ln|(√3u-1)/(√3u+1)| +C
由 sect=x+2,得 tant =√(x^2+4x+3),u=tan(t/2) = (x+1)/√(x^2+4x+3),
则 I = (1/√3)ln|[√3(x+1)-√(x^2+4x+3)]/[√3(x+1)+√(x^2+4x+3)]| +C.