求不定积分问题?

2025-02-27 10:42:54
推荐回答(1个)
回答1:

(1)
∫ x/√(2-3x^2) dx

=(-1/6)∫ d(2-3x^2)/√(2-3x^2)
=-(1/3) √(2-3x^2) + C

(2)
let
x= tanu
dx=(secu)^2 du
∫ xarctanx/(1+x^2)^(3/2) dx
=∫ [ u.tanu/(secu)^3] [(secu)^2 du]
=∫ u sinu du
=-∫ u dcosu
=-ucosu +∫ cosu du
=-ucosu +sinu + C
=-arctanx /√(1+x^2) + x/√(1+x^2) + C