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Đặt: $I=\int\sqrt{\tan x}dx$
Đặt: $\tan x=t\Rightarrow dt=(1+\tan^2x)dx=(1+t^2)dx$
Suy ra: $I=\int\frac{\sqrt t}{1+t^2}dt$
Đặt: $\sqrt t=s\Rightarrow t=s^2\Rightarrow dt=2sds$
Suy ra:
$I=2\int\frac{s^2}{s^4+1}ds$
$=2\int\left(\frac{s}{2\sqrt2(s^2-\sqrt2s+1)}-\frac{s}{2\sqrt2(s^2+\sqrt2s+1)}\right)ds$
$=\frac{1}{\sqrt2}\int\frac{s}{s^2-\sqrt2s+1}ds-\frac{1}{\sqrt2}\int\frac{s}{s^2+\sqrt2s+1}ds$
$=\frac{1}{2}\int\frac{1}{s^2-\sqrt2s+1}ds+\frac{1}{2\sqrt2}\int\frac{2s-\sqrt2}{s^2-\sqrt2s+1}ds+\frac{1}{2}\int\frac{1}{s^2+\sqrt2s+1}ds-\frac{1}{2\sqrt2}\int\frac{2s+\sqrt2}{s^2+\sqrt2s+1}ds$
$=\frac{1}{\sqrt2}\int\frac{d(\sqrt2s-1)}{(\sqrt2s-1)^2+1}+\frac{1}{2\sqrt2}\int\frac{d(s^2-\sqrt2s+1)}{s^2-\sqrt2s+1}+\frac{1}{\sqrt2}\int\frac{d(\sqrt2s+1)}{(\sqrt2s+1)^2+1}-\frac{1}{2\sqrt2}\int\frac{d(s^2+\sqrt2s+1)}{s^2+\sqrt2s+1}$
$=\frac{\arctan(\sqrt2s-1)}{\sqrt2}+\frac{\ln|s^2-\sqrt2s+1|}{2\sqrt2}+\frac{\arctan(\sqrt2s+1)}{\sqrt2}+\frac{\ln|s^2+\sqrt2s+1|}{2\sqrt2}+C$
$=\frac{\arctan(\sqrt{2t}-1)}{\sqrt2}+\frac{\ln|t-\sqrt{2t}+1|}{2\sqrt2}+\frac{\arctan(\sqrt{2t}+1)}{\sqrt2}+\frac{\ln|t+\sqrt{2t}+1|}{2\sqrt2}+C$
$=\frac{\arctan(\sqrt{2\tan x}-1)}{\sqrt2}+\frac{\ln|\tan x-\sqrt{2\tan x}+1|}{2\sqrt2}+\frac{\arctan(\sqrt{2\tan x}+1)}{\sqrt2}+\frac{\ln|\tan x+\sqrt{2\tan x}+1|}{2\sqrt2}+C$
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