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Câu hỏi: . Tìm nguyên hàm $I=\int{\dfrac{x}{\sqrt{2x+1}}dx}.$
A. $I=\dfrac{1}{6}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\dfrac{1}{2}\sqrt{2x+1}+C.$
B. $I=\dfrac{1}{6}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\sqrt{2x+1}+C.$
C. $I=\dfrac{1}{3}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\dfrac{1}{2}\sqrt{2x+1}+C.$
D. $I=\dfrac{1}{3}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\sqrt{2x+1}+C.$
A. $I=\dfrac{1}{6}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\dfrac{1}{2}\sqrt{2x+1}+C.$
B. $I=\dfrac{1}{6}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\sqrt{2x+1}+C.$
C. $I=\dfrac{1}{3}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\dfrac{1}{2}\sqrt{2x+1}+C.$
D. $I=\dfrac{1}{3}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\sqrt{2x+1}+C.$
Đặt $\sqrt{2x+1}=t\Rightarrow x=\dfrac{{{t}^{2}}-1}{2}\Rightarrow I=\int{\dfrac{\dfrac{{{t}^{2}}-1}{2}}{t}d\left( \dfrac{{{t}^{2}}-1}{2} \right)=\int{\dfrac{{{t}^{2}}-1}{2t}.tdt}=\dfrac{1}{2}\int{\left( {{t}^{2}}-1 \right)dt}}$
$=\dfrac{1}{2}\left( \dfrac{{{t}^{3}}}{3}-t \right)+C=\dfrac{{{t}^{3}}}{6}-\dfrac{t}{2}+C=\dfrac{1}{6}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\dfrac{1}{2}\sqrt{2x+1}+C.$
$=\dfrac{1}{2}\left( \dfrac{{{t}^{3}}}{3}-t \right)+C=\dfrac{{{t}^{3}}}{6}-\dfrac{t}{2}+C=\dfrac{1}{6}\sqrt{{{\left( 2x+1 \right)}^{3}}}-\dfrac{1}{2}\sqrt{2x+1}+C.$
Đáp án A.