Question

求微积分∫(x/(3+4x+x^2)^(1/2))dx

Answer 1

3+4x+x^2 = (x+2)^2 - 1,

(1) 当x>-1时,令 x+2 = sect,(3+4x+x^2)^(1/2) = tant

原式 = ∫ (sect-2) * sect dt = ∫ (sec²t - 2 sect) dt = tant - 2 ln|sect + tant| + C

= (3+4x+x^2)^(1/2) - 2 ln| x+2 + (3+4x+x^2)^(1/2) | + C

(2) 当x