Question

如图，求不定积分

Answer 1

换元,令t=√(1+x)/x=√(1+1/x),则t²=1+1/x,故x=1/(t²-1)
故∫1/x√(1+x)/xdx
=∫(t²-1)td(1/(t²-1))
=-∫(t²-1)t*2t/(t²-1)²dt
=-2∫t²/(t²-1)dt
=-2∫(1+1/(t²-1))dt
=-2∫(1+1/(t+1)(t-1))dt
而裂项公式1/(t+1)(t-1)=(1/(t-1)-1/(t+1))/2
所以
-2∫(1+1/(t+1)(t-1))dt
=-2∫dt-∫1/(t-1)dt+∫1/(t+1)dt
=-2t-ln绝对值(t-1)+ln绝对值(t+1)+C
(C为常数)
再把t=√(1+x)/x代入即可.

Answer 2

设 t = √[(1+x)/x] = √(1+1/x)，则 t² = 1 + 1/x， x = 1/(t²-1)，dx = -2t*dt/(t²-1)²
那么，上面的不定积分可以变换为：
=∫(t²-1) * t * (-2t * dt)/(t²-1)²
=∫-2t² * dt/(t²-1)
=-2*∫t²/(t²-1) * dt
=-2*∫[1 + 1/(t²-1)] *dt
=-1*∫[2 + 2/(t²-1)] * dt
=-1*∫2dt -∫[1/(t-1) - 1/(t+1)] * dt
=-2t - [∫dt/(t-1) -∫dt/(t+1)]
=-2t - [ln(t-1) - ln(t+1)] + C
=-2t - ln[(t-1)/(t+1)] + C
=-2t + ln[(t+1)/(t-1)] + C
把前面得到的 t = √(1+1/x) 代入上面的结果即可。