Question

不定积分：∫√(x+1)/x)dx

Answer 1

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]/[√(x+1)+1]| +C;

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