作业帮积分 根号下x/(根号下x+1)
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 07:25:00
积分 根号下x/(根号下x+1)
积分 根号下x/(根号下x+1)
积分 根号下x/(根号下x+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
I = integral 1/sqrt(x^2+x) dx
= integral 1/sqrt((x+1/2)^2-1/4) dx
substitute u = x+1/2 and du = dx:
I= integral 1/sqrt(u^2-1/4) du
substitute u = (sec(s))/2 and du = 1/2...
全部展开
I = integral 1/sqrt(x^2+x) dx
= integral 1/sqrt((x+1/2)^2-1/4) dx
substitute u = x+1/2 and du = dx:
I= integral 1/sqrt(u^2-1/4) du
substitute u = (sec(s))/2 and du = 1/2 tan(s) sec(s) ds.
Then sqrt(u^2-1/4) = sqrt((sec^2(s))/4-1/4) = (tan(s))/2 and s = sec^(-1)(2 u):
I= integral sec(s) ds
= ln(tan(s)+sec(s))+constant
Substitute back for s = sec^(-1)(2 u):
I = ln(sqrt(4 u^2-1)+2 u)+constant
Substitute back for u = x+1/2:
I = ln(2 x+2 sqrt(x (x+1))+1)+constant
= ln( x+2 sqrt(x (x+1))+(x+1) )+constant
=2 ln ( sqrt(x) + sqrt(x+1) ) + constant
收起