高数不定积分问题求解答 1/x（根号下x的平方-1） dx 如果我令x=sect那么 d
没错啊，积分结果出来不就是t+c,
然后你得把t换成x啊。sect=x，得到t=arccos(1/x)
根号下[x/（1-x）]不定积分
∫√(x/(1-x))) dx
= ∫ √[1/(1-x) -1] dx
let
1/(1-x) = (seca)^2
[1/(1-x)^2] dx = 2(seca)^2tana da
dx = 2[tana/(seca)^2] da
∫ √[1/(1-x) -1] dx
=∫ tana (2tana/(seca)^2) da
=2∫ (sina)^2 da
=∫ (1-cos2a) da
=a- sin(2a)/2 + C
= arccos(√(1-x)) -√x√(1-x) + C
x乘根号下x 1的不定积分
∫x.√(x+1) dx
=∫(x+1)^(3/2) dx - ∫√(x+1) dx
=(2/5)(x+1)^(5/2) - (2/3)(x+1)^(3/2) + C
(1/x)*根号下(x+1)/(x-1) dx？
令 √[(x+1)/(x-1)] = u, x+1 = u^2(x-1), (u^2-1)x = 1+u^2,
x = (u^2+1)/(u^2-1) = 1 + 2/(u^2-1), dx = -4udu/(u^2-1)^2
I = ∫(1/x)√[(x+1)/(x-1)]dx = ∫[(u^2-1)/(u^2+1)]u[-4udu/(u^2-1)^2]
= -4∫u^2du/[(u^2+1)(u^2-1)] = -2∫[1/(u^2+1) + 1/(u^2-1)]du
= -2∫du/(u^2+1) - ∫[1/(u-1)-1/(u+1)]du
= -2arctanu - ln|(u-1)/(u+1)| + C
= -2arctan√[(x+1)/(x-1)] - ln|(√[(x+1)/(x-1)]-1)/(√[(x+1)/(x-1)]+1)| + C