∫1/(1+2根号x)dx

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∫1/(1+2根号x)dx∫1/(1+2根号x)dx∫1/(1+2根号x)dx令t=√x∫1/(1+2√x)dx=∫1/(1+2t)dt^2=∫2t/(1+2t)dt=∫1-1/(1+2t)dt=∫d

∫1/(1+2根号x)dx
∫1/(1+2根号x)dx

∫1/(1+2根号x)dx
令 t = √x
∫1/(1+2√x)dx = ∫1/(1+2t)dt^2 = ∫ 2t/(1+2t)dt = ∫ 1- 1/(1+2t)dt = ∫ dt - ∫1/(1+2t)dt
= t + 1/2 ln(1+2t) + C
= √x + 1/2 ln(1+2√x) + C

令2√X=t,则4x=t^,dx=dt^/4
则,原式=∫1/(1+t)dt^/4
=∫(1/4)2t/1+tdt
=1/2 ∫ (t+1-1)/1+tdt
=1/2 ∫ 1dt - 1/2 ∫ 1/1+tdt
=(1/2)t -1/2ln(1+t) +C
还原得:原式=√x - 1/2 ln(1+2√x) + C

解法:

令 t = √x
∫1/(1+2√x)dx = ∫1/(1+2t)dt^2
= ∫ 2t/(1+2t)dt = ∫ 1- 1/(1+2t)dt = ∫ dt - ∫1/(1+2t)dt
= t + 1/2 ln(1+2t) + C
= √x + 1/2 ln(1+2√x) + C
这种做法不错!!