不定积分 ∫dx⼀［（x-a）（x-b）］^봀 其中a＜b是常数 求具体过程步骤重点是过程不是答

√(b - x) = √[b - a - (x - a)] = √[b - a - √(x - a)²]
1/√(x - a) dx = 2 · 1/[2√(x - a)] d(x - a) = 2 d√(x - a)

我的做法：
∫ dx/√[(x - a)(x - b)] = ∫ dx/√[x² - (a + b)x + ab]
= ∫ dx/√[(x - (a + b)/2)² - ((a + b)/2)² + ab]
= ∫ dx/√[(x - (a + b)/2)² - ((a - b)/2)²]
= ln|(a - b)/2 + √[(x - (a + b)/2)² - ((a - b)/2)²]| + C <==∫ dx/√(x² - a²) = ln|x + √(x² - a²)| + C
= ln|(a - b)/2 + √(x - a)(x - b)| + C
= ln|√(a - x)² + √(x - b)² + 2√(x - a)(x - b)| + C''，C'' = C - ln2
= ln|[√(x - a) + √(x - b)]²| + C''
= 2ln|√(x - a) + √(x - b)| + C''

被积函数到底是啥？图呢？

1如果题目：
∫dx/√(x^2-m^2)
x=msecu dx=msecutanudu tanu=√[(x/m)^2-1]
=∫secudu
=∫du/cosu
=(1/2)ln|1+sinu|/|1-sinu|+C
=ln|1+sinu|/|cosu|+C
=ln|secu+tanu|+C
=ln|x/m+√(x/m)^2-1|+C
m>0
=ln|x+√(x^2-m^2)|+C1
m<0
=ln|x-√(x^2-m^2)|+C1

∫dx/[√(x-a)(x-b)]=∫dx/√[[x-(a+b)/2]^2 -(a+b)^2/4+ab]
=∫dx/√[(x-(a+b)/2)^2-(a-b)^2/4]
m=(a-b)/2
(a-b)>0时
=ln|[x-(a+b)/2] +√(x-a)(x-b)|+C
a-b<0时
=ln|[x-(a+b)/2] -√(x-a)(x-b)|+C

2题目
∫dx/√(m^2-x^2)
=arcsin(x/|m|)+C
m>0
=∫d(x/m)/√(1-x^2/m^2)
=arcsin(x/m)+C
m<0
=∫d(-x/m)/√(1-x^2/m^2)
=arcsin(-x/m)+C

∫dx/√(x-a)(b-x)
=∫dx/√[-x^2+(a+b)x-ab]
=∫dx/√[(a-b)^2/4 -[x-(a+b)/2]^2]
=arcsin[ (x-(a+b)/2)/|(a-b)/2| ]+C
=arcsin [ (2x-a-b)/|a-b| ] +C