Question

求x^4/(x+3)dx的不定积分

Answer 1

∫ x^4/(x+3) dx
=∫ (x+3-3)^4/(x+3) dx
=∫ [(x+3)^4+4(-3)(x+3)^3+6(-3)^2(x+3)^2+4(-3)^3(x+3)+(-3)^4]/(x+3) dx
=∫ [(x+3)^3 - 12(x+3)^2 + 54(x+3) -108 + 81/(x+3)] dx
=∫ (x+3)^3 dx - 12∫ (x+3)^2 dx + 54∫ (x+3) dx - 108∫ 1 dx +81∫ 1/(x+3) dx
=(x+3)^4/4 - 12(x+3)^3/3 + 54(x+3)^2/2 - 108x +81ln|x+3| + C
=(x+3)^4/4 - 4(x+3)^3 + 27(x+3)^2 - 108x +81ln|x+3| + C
=x^4/4 - x^3 + 9x^2/2 - 27x + 81ln|x+3| + C
有不懂欢迎追问

Answer 2

原式=∫(x³-3x²+9x-27+81/(x+3)]dx
=x^4/3-x³+9x²/2-27x+81ln|x+3|+C

Answer 3

设t=x+3,则有dt=dx
原式即为求取(t-3)^4/t*dt,需要用2项式定理展开下(t-3)^4,在逐个求取即可.

Answer 4

x^4/(x+3)=x^3-3x^2+9x-27+81/(x+3)

然后会积了吧？