求不定积分 ∫(2-x+x^3)dx

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

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

∫(x+1-3)/(x²+2x+3)dx
=∫(x+1)/(x²+2x+3)dx-∫3/(x²+2x+3)dx
=(1/2)∫1/(x²+2x+3)d(x²+2x+3)-∫3/(x²+2x+3)dx
=(1/2)ln(x²+2x+3)
-
∫3/[(x+1)²+2]dx
=(1/2)ln(x²+2x+3)
-
(1/2)∫3/[(1/√2x+1/√2)²+1]dx
=(1/2)ln(x²+2x+3)
-
(3√2/2)∫1/[(x+1)²+2]d(1/√2
x+1/√2)
=(1/2)ln(x²+2x+3)
-
(3√2/2)arctan(1/√2
x+1/√2)+C

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

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