Question

求不定积分∫x^3/根号下(9+X^2) dx 需要过程

Answer 1

x=3tana
x²+9=9sec²a
seca=√(x²+9)/3
dx=3sec²ada
原式=∫(27tan³a)*3sec²ada/3seca
=∫(27tan³aseca)da
=27∫tan²adseca
=27∫(sec²a-1)dseca
=9sec³a-27seca+C
=(x²+9)√(x²+9)/3-9√(x²+9)+C

Answer 2

∫x^3dx/√(9+x^2)
=(1/2)∫x^2dx^2/√(9+x^2)
=(1/2)∫∫√(9+x^2)dx^2-(9/2)∫dx^2/√(9+x^2)
=(1/3)√(9+x^2)^3 -9√(x^2+9) +C

追问

两个积分号叠加在一起是什么意思啊?你确定你没有犯什么错误吗？能不能把步骤写详细点？

追答

报歉，多复制了一个积分号，其它没有错误
√(9+x^2)^3 =(9+x^2)^(3/2)=(9+x^2)√(9+x^2)

∫x^3dx/(9+x^2)^(1/2)
=(1/2)∫x^2dx^2/(9+x^2)^(1/2)
=(1/2)∫(9+x^2)^(1/2)dx^2-(9/2)∫dx^2/(9+x^2)^(1/2)
=(1/3)(9+x^2)^(3/2) -9(x^2+9)^(1/2) +C