[Toan12] Tính tích phân

V

vivietnam

Đặt $ \sqrt{2x+1}=t \Longrightarrow x=\dfrac{t^2-1}{2} \Longrightarrow dx=tdt $
$ I= \displaystyle \int \dfrac{t^2dt}{1+t}=\displaystyle \int (t-1+\dfrac{1}{t+1})dt=\dfrac{t^2}{2}-t+ln|t+1|$
 
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