不定积分∫[√（1+x^2)]/xdx=?我想看下步骤,

Question

不定积分∫[√（1+x^2)]/xdx=?我想看下步骤,

Answer 1

∫ sqrt(1+x^2)/x dx
令 x = tan(u) 则 dx = sec^2(u) du. sqrt(x^2+1) = sqrt(tan^2(u)+1) = sec(u) , u = tan^(-1)(x):
= ∫ csc(u) sec^2(u) du
= ∫ (tan^2(u)+1) csc(u) du
= ∫ (csc(u)+tan(u) sec(u)) du
= ∫ csc(u) du+ ∫ tan(u) sec(u) du
令s = sec(u) 则 ds = tan(u) sec(u) du:
= ∫ 1 ds+ ∫ csc(u) du
= ∫ 1 ds-ln(cot(u)+csc(u))
= s-ln(cot(u)+csc(u))+C
代回 s = sec(u):
= sec(u)-ln(cot(u)+csc(u))+C
代回 u = tan^(-1)(x):
= sqrt(x^2+1)-ln((sqrt(x^2+1)+1)/x)+C
= sqrt(x^2+1)-ln(sqrt(x^2+1)+1)+ln(x)+C