Question

高数求不定积分 总习题四17

往下该怎么算？
如图

Answer 1

下来原式=-(1/2)∫[(u^2)/√(1+u^2)]d(u^2)
=-(1/2)∫{[(1+u^2)-1]/√(1+u^2)}d(u^2)
=-(1/2)∫√(1+u^2)d(1+u^2)+(1/2)∫[1/√(1+u^2)]d(u^2)
=-(1/3)*(1+u^2)^(3/2)+(1/2)arctanu+c
再将u=1/x代入得
-(1/3)*(1+1/x^2)^(3/2)+(1/2)arctan(1/x)+c

Answer 2

继续，设 v = √(1+u²), v² = 1+u², v dv = u du
I = ∫ ﹣u² u /√(1+u²) du = ﹣∫ (v²-1) dv
= (﹣1/3)v³ + v + C
= (-1/3) (x²+1)^(3/2) / x³ + √(x²+1)/x + C

利用 u³ = u * u² = u [(1+u²) ﹣1] 即得。

Answer 3

可以分部积分：udu/√(1+u^2)=d√(1+u^2)，原式=-∫u^2d√(1+u^2)，分部后可得到结果