求不定积分∫x^3⼀根号下(9+X^2) dx 需要过程

2024-12-27 17:26:55
推荐回答(2个)
回答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

回答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