Question

求定积分∫₀¹(χ²+3χ+2)÷χdχ

Answer 1

反常积分啦

Answer 2

上下同时除以√x，
=xsin(1/x)/√(1+1/x) ，令1/x=t
=sint/t*1/√(1+t)
t→0,sint~t
=1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
上下同时除以x，
=xsin(1/2x)/√(1-2/x) ，令1/x=t
=sin(t/2)/t*1/√(1-2t)
t→0,sint~t
=lim(t/2)/t
=1/2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
x→0-时，|x|=-x
则原式=limsinx/(-x)
x→0,sinx~x
原式=-1

Answer 3

∫[(x-3)/(x^2-x+1 )]dx

=1/2∫[(2x-1-5)/(x^2-x+1 )]dx

=1/2∫[(2x-1)/(x^2-x+1 )]d(x^2-x+1)-5/2∫/(x^2-x+1 )dx
=1/2ln(x^2-x+1 )-5/2∫/(x^2-x+1 )dx
=1/2ln(x^2-x+1 )-(5√3)/3*arctan[√3*(2x-1)/3]+C
其中：
5/2∫dx/x^2-x+1
=5/2∫dx/[(x-1/2)^2+3/4]
=5/2∫d(x-1/2)/[(x-1/2)^2+3/4]
=5/2*(2/√3)*arctan[(x-1/2)/(√3/2)]+C
=(5√3)/3*arctan[√3*(2x-1)/3]+C