∫1/x√(2x-1)dx

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/17 09:29:17

∫1/x√(2x-1)dx

∫dx/[x√(2x-1)]
let
x= (1/2) (secy)^2
dx = (secy)^2.(tany) dy
∫dx/[x√(2x-1)]
=2∫ dy
=2y + C
=2arccos (1/√(2x)) + C
再问：专科生理解不了
再问：专科生理解不了。答案是2arctan√(2x-1)+C
再答： ∫dx/[x√(2x-1)]
let

x= (1/2) (secy)^2
dx = (secy)^2. (tany) dy

∫dx/[x√(2x-1)]

=∫ (secy)^2. (tany) dy / [(1/2) (secy)^2 . √ (secy^2 -1) ]
=∫ (secy)^2. (tany) dy / [(1/2) (secy)^2 . tany ]
=2∫ dy
=2y + C
=2arccos (1/√(2x)) + C
=2arctan√(2x-1) + C

∫x√(1+2x)dx 求不定积分 ∫ 1/(1+2x)² dx ∫ x/√x²+4 dx ∫dx/x+√(1-x²) ∫1/√x*(4-x)dx ∫dx/[x√(1-x^4)] ∫1/(x^2+x+1)dx ∫X^2/1-x^2 dx. ∫dx/x^2(1-x^2) ∫dx/x^2(1+x^2) ∫（x^2+1/x^4)dx ∫(1-x)^2/x^3 dx ∫ x/(1+X^2)dx=