求不定积分 ∫(2-x+x^3)dx

因
∫2dx = 2x +C
∫xdx = (1/2)x^2 +C
∫x^3dx = (1/4)x^4 +C

所以
∫(2-x+x^3)dx

= ∫2dx - ∫xdx +∫x^3dx

= （2x +C1) - [(1/2)x^2 +C2] + [(1/4)x^4 +C3]

= (1/4)x^4 - (1/2)x^2 + 2x + C

2x-1/2x^2+1/4x^4+c

=2x-x^2/2+x^4/4+c

=2x-x^2/2+x^4/4+c

首先第1步=∫(2-x)dx+∫(x^3)dx
=∫2dx-∫xdx+∫(x^3)dx
=2x+c-(x^2/2 + c)+(x^4/4+c)
由于C代表一个常数,所以c-c+c还是一个常数,还可以用C表示
=2x - x^2/2 + x^4/4 + c