Question

高数，不定积分，ʃ(x+3)/√(-4x-x^2) dx=?

如图

Answer 1

I = ʃ [(x+3)/√(-4x-x^2)] dx = ʃ [(x+3)/√[4-(x+2)^2] dx
令 x+2 = 2sint， 则 dx = 2costdt
I = ʃ [(2sint + 1)/(2cost)] 2costdt = ʃ (2sint + 1)dt
= -2cost + t + C = √(-4x-x^2) + arcsin[(x+2)/2] + C

Answer 2

-4x-x^2

= 4-(x^2+4x+4)
=4-(x+2)^2
let
x+2 = 2sinu
dx=2cosu du
∫ (x+3)/√(-4x-x^2) dx

=∫ (2sinu +1) du
=-2cosu +u +C
=-√(-4x-x^2) + arcsin[(x+2)/2] +C

Answer 3