∫
___.
int dfrac (1)({x)^2}sqrt (dfrac {1-x)(1+x)}dx
不定积分int dfrac (1)(1+sqrt {1-x)}dx=( ) int dfrac (1)(1+sqrt {1-x)}dx=int dfrac (
int dfrac (1-x)(sqrt {9-4{x)^2}}dx=________=________
(int )_(0)^1dfrac (dx)({x)^2sqrt (1-x)} 发散-|||-D (int )_(0)^1dfrac (dx)(xsqrt {1
(26) int dfrac (1-x)(sqrt {9-4{x)^2}}dx
设=ln sqrt (dfrac {1-x)(1-{x)^2}}则 dy|=ln sqrt (dfrac {1-x)(1-{x)^2}}设则dy|
(int )_(dfrac {3)(4)}^1dfrac (dx)(sqrt {1-x)-1}..
=dfrac (sqrt {1+x)-sqrt (1-x)}(sqrt {1+x)+sqrt (1-x)}
(9) =dfrac (sqrt {1+x)-sqrt (1-x)}(sqrt {1+x)+sqrt (1-x)} ;
(6) int_((1)/(sqrt(2)))^1(sqrt(1-x^2))/(x^2)dx;(6) $\int_{\frac{1}{\sqrt{2}}}^{1