Question

∫xdx/根号下（1+x-x^2）,怎么下手。。。

Answer 1

∫ x/√(1 + x - x²) dx
= ∫ x/√[5/4 - (x - 1/2)²] dx
Let x - 1/2 = √5/2 siny and dx = √5/2 cosy dy
Then √[5/4 - (x - 1/2)²] = √[5/4 - 5/4 sin²y] = √5/2 cosy
x = √5/2 siny + 1/2
==> ∫ (√5/2 siny + 1/2)/(√5/2 cosy) * (√5/2 cosy) dy
= (√5/2)∫ siny dy + (1/2)∫ dy
= (√5/2)(-cosy) + (1/2)y + C
= (-√5/2) * 2√(1 + x - x²)/√5 + (1/2) arcsin[(2x - 1)/√5] + C
= -√(1 + x - x²) + (1/2) arcsin[(2x - 1)/√5] + C
Notes：
siny = (x - 1/2)/(√5/2) = (2x - 1)/√5
cosy = √[5 - (2x - 1)²]/√5 = 2√(1 + x - x²)/√5