求不定积分∫xdx⼀√(x-3)和∫√(x^2-9)dx⼀x

2025-06-28 09:04:12
推荐回答(1个)
回答1:

1、原式=∫(x-3+3)dx/√(x-3)
=∫√(x-3)dx+∫3dx/√(x-3)
=∫√(x-3)d(x-3)+3∫d(x-3)/√(x-3)
=(x-3)^(1/2+1)/(1/2+1)+3(x-3)^(-1/2+1)/(-1/2+1)+C
=(2/3)(x-3)^(3/2)+6(x-3)^(1/2)+C.
2、设x=3sect,
(sect)^2=x^2/9,
(tant)^2=(x^2-9)/9,
t=arcsin[√(x^2-9)/x]
dx=3secttantdt,
x^2-9=9(tant)^2,
原式=∫3tant*(3sect*tant)dt/(3sect)
=3∫(tant)^2dt
=3∫[(sect)^2-1]dt
=3(tant-t)+C
=√(x^2-9)-3arcsin[√(x^2-9)/x]+C