设 (int )_(0)^xuf(u)du=(e)^x-1, 则 (int )_(0)^sqrt (ln 2)(x)^3f((x)^2)dx= ()-|||-(A)1 (B) -1-|||-(C) dfrac (1)(2) (D) dfrac (1)(2)e

设(x)=dfrac (1)(1+{x)^2}+sqrt (1-{x)^2}(int )_(0)^1f(x)dx, 则 (int )_(0)^1f(x)dx=设

(5) (int )_(0)^ln 2sqrt ({e)^x-1}dx

(8) (int )_(0)^1(x)^2sqrt (1-{x)^2}dx :

设 gt 0, 则 =(int )_(-a)^asqrt ({a)^2-(x)^2}ln dfrac (x+sqrt {1+{x)^2}}(3)dx= __

(3)若 int f(x)dx=(x)^2+C, 则 int xf(1-(x)^2)dx= __ ;

求不定积分int dfrac (1)(2x)sqrt (ln x)dx=（）.int dfrac (1)(2x)sqrt (ln x)dx=int dfrac

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

[2023年真题]设连续函数f(x)满足： f(x+2)-f(x)=x,int_(0)^2f(x)dx=0,则 int_(1)^3f(x)dx=[2023年真题

计算(int )_(0)^ln 2sqrt ({e)^x-1}dx时，常用换元法，若令(int )_(0)^ln 2sqrt ({e)^x-1}dx，则换元后定

(int )_(1)^(e^2)dfrac (ln x)(sqrt {x)}dx= __