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

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

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

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

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

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

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

求(int )_(1)^+infty dfrac (dx)(xsqrt {2{x)^2-1}}=求

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

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

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