5【判断题】 判断:反常积分int_(0)^(1)/(2)(1)/(sqrt(x(1-x)))dx是发散的.（）

2.[判断题] 判断：int_(0)^1x^m(1-x)^ndx=int_(0)^1x^n(1-x)^mdx,(m,nin N).（）A. 对B. 错

(int )_(0)^1dfrac (dx)({x)^2sqrt (1-x)} 发散-|||-D (int )_(0)^1dfrac (dx)(xsqrt {1

(6) int_((1)/(sqrt(2)))^1(sqrt(1-x^2))/(x^2)dx;(6) $\int_{\frac{1}{\sqrt{2}}}^{1

定积分 int_(0)^1 (1)/(1 + sqrt(x)) dx = ( )A. $1 - 2\ln 2$B. $2 - \ln 2$C. $2 - 2\l

不定积分int dfrac (1)(1+sqrt {1-x)}dx=( ) int dfrac (1)(1+sqrt {1-x)}dx=int dfrac (

int_(0)^2(1)/(1+sqrt[3](x))dx;$\int_{0}^{2}\frac{1}{1+\sqrt[3]{x}}dx;$

定积分(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0

求定积分int_(-1)^1 ((xrm{dx))/(sqrt(5-4x)) }求定积分$\int_{-1}^{1} {\frac{x\rm{dx}}{\sqr

2.计算定积分int_(0)^1ln(1+sqrt(x))dx.2.计算定积分$\int_{0}^{1}\ln(1+\sqrt{x})dx$.

积分 int_(0)^1 xe^2x , dx = ____.A. $\frac{e^2+1}{4}$B. $1$C. $\frac{e^2+1}{2}$D.