I = ∫√(1+x)dx/[x√(1-x)] = ∫√(1-x^2)dx/[x(1-x)] (设 x = sint)
= ∫(cost)^2dt/[sint(1-sint)]
= ∫[1-(sint)^2][1/sint+1/(1-sint)]dt
= ∫[csct-sint + 1+sint]dt = ∫(1+csct)dt
= t + ln|csct - cott| + C
= arcsinx + ln| [1-√(1-x)]/x | + C
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024