Question

积分 根号下x/(根号下x+1)

Answer 1

答：
设t=√[x/(x+1)]
t^2=(x+1-1)/(x+1)=1-1/(x+1)
1/(x+1)=1-t^2
x+1=1/(1-t^2)
x=-1+1/[(1-t)(1+t)]
x=-1+(1/2)*[ 1/(1-t)+1/(1+t) ]
∫ √x /√(x+1) dx
=∫ t *(1/2) d [1/(1-t) +1/(1+t) ]
=(1/2) ∫ t*[1/(1-t)^2-1/(1+t)^2] dt
=(1/2) ∫ -(1-t-1) / (1-t)^2 dt -(1/2)∫ (1+t-1) /(1+t)^2 dt
=(1/2) ∫ -1/(1-t) +1/(1-t)^2 dt -(1/2) ∫ 1/(1+t) -1/(1+t)^2 dt
=(1/2)*ln(t-1) +(1/2) / (t-1) -(1/2)*ln(t+1) -(1/2)/(t+1)+C
=(1/2) ln[(t-1)/(t+1) ]+1/ (t^2-1)+C
t=√[x/(x+1)]代入即可