t = arctan√x ,sect = √(1+x),x = tan² t ,dx = 2 tan t * sec² t dt原式 = ∫ 2 t d(sect) = 2 t * sect - 2∫sect dt= 2 t * sect - 2 ln|sect + tant| + C= 2 √(1+x) arctan√x - 2 ln|√(1+x) +√x...
求 ∫[arctan√x/√(1+x)]dx 的不定积分.√表示根号,
t = arctan√x ,sect = √(1+x),x = tan² t ,dx = 2 tan t * sec² t dt原式 = ∫ 2 t d(sect) = 2 t * sect - 2∫sect dt= 2 t * sect - 2 ln|sect + tant| + C= 2 √(1+x) arctan√x - 2 ln|√(1+x) +√x...