int dfrac (1)(sqrt {{x)^2-1}}dx 时应该令 int dfrac (1)(sqrt {{x)^2-1}}dx以化去根在计算式 A 对 B 错

(int )_(1)^2dfrac (sqrt {{x)^2-1}}(x)dx

不定积分 int dfrac ({x)^2-1}(sqrt {xsqrt {x)}}dx=

)int dfrac (dx)(2{x)^2-1};

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

计算下列定积分.-|||-(int )_(sqrt {2)}^2dfrac (dx)(xsqrt {{x)^2-1}} ;

int dfrac (dx)(xsqrt {{x)^2-1}}=____________________________

(int )_(1)^2dfrac (dx)(x(1+sqrt {2x))}=( )A.(int )_(1)^2dfrac (dx)(x(1+sqrt {2x

(9) int dfrac (1)(xsqrt {{x)^2-1}}dx

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

(37 ) int dfrac (dx)(xsqrt {{x)^2-1}} ,