求不定积分 ∫1⼀(x+x^2)^(1⼀2)

2024-12-17 05:04:18

推荐回答（1个）

回答1：

∫1/(x+x^2)^(1/2）dx= ∫1/[(x+1/2)^2-1/4]^(1/2)dx

当x+1/2>1/2时令 x+1/2=1/2sect ( 0则 ∫1/[(x+1/2)^2-1/4]^(1/2)dx
=∫1/[1/4sec^2t-1/4]^(1/2)d(1/2sect-1/2)
=∫1/(1/4tan^2t)^(1/2) *(1/2)* sect tant dt
=∫sectdt=ln(sect+tant) +C
=ln{x+1/2+[(x+1/2)^2-1/4]^(1/2)}+C1

当x+1/2<-1/2时令 x+1/2=-u 得u>1/2 u符合上式条件将u代人上式
得∫1/[(x+1/2)^2-1/4]^(1/2)dx
=ln{u+[(u^2-1/4]^(1/2)}+C1
=ln{-x-1/2-[(x+1/2)^2-1/4]^(1/2)}+C1

综上所述 ∫1/(x+x^2)^(1/2)=lnl x+1/2+[(x+1/2)^2-1/4]^(1/2) l+C