Question

不定积分∫(x-1)/√(1-4x^2)dx

Answer 1

因为分母是√1-4x^2
所以设 x=cost/2
dx=-sint/2 *dt
则√1-4x^2=sint
所以原式化为
∫(cost/2-1)/sint *(-sint/2 *dt)
=1/2∫(1-cost/2)dt
=1/2∫dt -1/4 * ∫costdt
=1/2*t-1/4∫dsint
=t/2-sint/4
因为
x=cost/2
t=arccos(2x)
所以
∫(cost/2-1)/sint *(-sint/2 *dt)
=arccos(2x)/2-sinarccos(2x)/4
=arccos(2x)/2-√(1-cos^2(cos(2x))/4
=arccos(2x)/2-√(1-4x^2) / 4 (注意分母4在根号外)

Answer 2

∫(x-1)/√(1-4x^2)dx
= (-1/4) ∫d√(1-4x^2) - ∫1/√(1-4x^2)dx
=(-1/4) √(1-4x^2) - ∫1/√(1-4x^2)dx
let
2x= sina
2dx= cosada
∫1/√(1-4x^2)dx
=(1/4)∫da
= a/4 + C
=arcsin(2x) + C'

∫(x-1)/√(1-4x^2)dx
=(-1/4) √(1-4x^2) - ∫1/√(1-4x^2)dx
=(-1/4) √(1-4x^2) - arcsin(2x) + C