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Đặt $\arctan x=t\Rightarrow x=\tan t$
$\Rightarrow dx=(1+\tan^2t)dt$
Từ đó:
$\int h(x)dx=\int\frac{e^t\tan t(1+\tan^2t)dt}{(1+\tan^2t)^{\frac{3}{2}}}$
$=\int\frac{e^t\tan tdt}{\sqrt{1+\tan^2t}}$
$=\int e^t\sin tdt$
$=\frac{(\sin t-\cos t)e^t}{2}+C$
$=\frac{(\tan t-1)e^t}{2\sqrt{1+\tan^2t}}+C$
$=\frac{(x-1)e^{\arctan x}}{2\sqrt{1+x^2}}+C$
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