不定积分：∫√(x+1)/x)dx

若是 I = ∫ [√(x+1)/x] dx,令√(x+1) = t,则 x=t^2-1,I = ∫ [√(x+1)/x] dx = ∫ 2t^2dt/(t^2-1) = 2 ∫ [1+1/(t^2-1)]dt= 2t+ ∫ [1/(t-1)-1/(t+1)]dt = 2t+ ln|(t-1)/(t+1)| +C= 2√(x+1)+ ln|[√(x+1)-1]/[√(...