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Câu hỏi: Họ các nguyên hàm của hàm số $f\left( x \right)=\dfrac{x}{\sqrt{2x+1}}$ trên khoảng $\left( -\dfrac{1}{2}; +\infty \right)$ là
A. $\dfrac{1}{3}\left( x-2 \right)\sqrt{2x+1}+C$.
B. $\dfrac{1}{3}\left( x-1 \right)\sqrt{2x+1}+C$.
C. $\dfrac{1}{3}x\sqrt{2x+1}+C$.
D. $\dfrac{1}{6}\left( 2x-1 \right)\sqrt{2x+1}+C$.
A. $\dfrac{1}{3}\left( x-2 \right)\sqrt{2x+1}+C$.
B. $\dfrac{1}{3}\left( x-1 \right)\sqrt{2x+1}+C$.
C. $\dfrac{1}{3}x\sqrt{2x+1}+C$.
D. $\dfrac{1}{6}\left( 2x-1 \right)\sqrt{2x+1}+C$.
Ta có $I=\int{\dfrac{x}{\sqrt{2x+1}} }dx$. Đặt $t=\sqrt{2x+1}\Rightarrow {{t}^{2}}=2x+1\Rightarrow tdt=dx$.
Khi đó $I=\int{\dfrac{{{t}^{2}}-1}{2t}}.tdt=\dfrac{1}{2}\int{\left( {{t}^{2}}-1 \right)dt=\dfrac{1}{2}}\left( \dfrac{{{t}^{3}}}{3}-t \right)+C=\dfrac{1}{2}t\left( \dfrac{{{t}^{2}}}{3}-1 \right)+C$
$=\dfrac{1}{2}\sqrt{2x+1}\left( \dfrac{2x+1}{3}-1 \right)+C=\dfrac{1}{3}\sqrt{2x+1}\left( x-1 \right)+C$.
Khi đó $I=\int{\dfrac{{{t}^{2}}-1}{2t}}.tdt=\dfrac{1}{2}\int{\left( {{t}^{2}}-1 \right)dt=\dfrac{1}{2}}\left( \dfrac{{{t}^{3}}}{3}-t \right)+C=\dfrac{1}{2}t\left( \dfrac{{{t}^{2}}}{3}-1 \right)+C$
$=\dfrac{1}{2}\sqrt{2x+1}\left( \dfrac{2x+1}{3}-1 \right)+C=\dfrac{1}{3}\sqrt{2x+1}\left( x-1 \right)+C$.
Đáp án B.