[Toán 12] Tính nguyên hàm

T

talathangngoc

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V

vivietnam

$I=\int \dfrac{\sqrt{1+\sqrt{x}}}{x}dx$
Đặt $\sqrt{x}=t \Longrightarrow dx=2tdt $
$I=\int \dfrac{\sqrt{1+t}.2tdt}{t^2}=\int \dfrac{2\sqrt{1+t}dt}{t}$
Đặt $\sqrt{1+t}=u \Longrightarrow dt=2udu$
$I=\int \dfrac{2u.2udu}{u^2-1}=\int (4+\dfrac{4}{u^2-1})du=\int (4+\dfrac{2}{u-1}-\dfrac{2}{u+1})du=4u+2ln|\dfrac{u-1}{u+1}|+C $
 
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