x^4⼀(x+1)的不定积分

解：
x^4/(x+1)=x³-x²+x+1/(x+1)-1
∫x^4/(x+1) dx
=∫[x³-x²+x+1/(x+1)-1]dx
=∫x³dx-∫x²dx+∫xdx+∫1/(x+1)-∫1dx
=x^4/4-x³/3+x²/2+ln|x+1|-x+C

x^4/(x+1)
=[(x+1)-1]^4/(x+1)
=[(x+1)^4-4(x+1)^3+6(x+1)^2-4(x+1)+1]/(x+1)
=(x+1)^3-4(x+1)^2+6(x+1)-4+1/(x+1)
原式=∫[(x+1)^3-4(x+1)^2+6(x+1)-4+1/(x+1)]dx
=1/4*(x+1)^4-4/3*(x+1)^3+3(x+1)^2-4(x+1)+ln|x+1|+C

x⁴/(x+1)=(x⁴-1+1)/(x+1)
=[(x²-1)(x²+1)]/(x+1) +1/(x+1)
=(x-1)(x+1)(x²+1)/(x+1) +1/(x+1)
=(x-1)(x²+1) +1/(x+1)
=x³-x²+x-1 +1/(x+1)

∫x⁴/(x+1) dx
=∫[x³-x²+x-1 +1/(x+1)]dx
=x⁴/4 -x³/3 +x/2 -x +ln(x+1)+C