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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)=\sqrt{2x-1}$ thỏa mãn $F\left( 1 \right)=\dfrac{4}{3}.$ Tìm $F\left( x \right).$
A. $F\left( x \right)=-\dfrac{1}{3}\sqrt{2x-1}+\dfrac{5}{3}.$
B. $F\left( x \right)=\dfrac{1}{3}\sqrt{2x-1}+1.$
C. $F\left( x \right)=-\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}+\dfrac{5}{3}.$
D. $F\left( x \right)=\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}+1.$
A. $F\left( x \right)=-\dfrac{1}{3}\sqrt{2x-1}+\dfrac{5}{3}.$
B. $F\left( x \right)=\dfrac{1}{3}\sqrt{2x-1}+1.$
C. $F\left( x \right)=-\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}+\dfrac{5}{3}.$
D. $F\left( x \right)=\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}+1.$
Ta có $I=F\left( x \right)=\int{\sqrt{2x-1}}dx.$
Đặt $t=\sqrt{2x-1}\Rightarrow I=\int{td\left( \dfrac{{{t}^{2}}+1}{2} \right)}=\int{t.tdt}=\dfrac{{{t}^{3}}}{3}+C\Rightarrow F\left( x \right)=\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}+C.$
Mà $F\left( 1 \right)=\dfrac{4}{3}\Rightarrow \dfrac{1}{3}+C=\dfrac{4}{3}\Rightarrow C=1\Rightarrow F\left( x \right)=\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{2}}}+1.$
Đặt $t=\sqrt{2x-1}\Rightarrow I=\int{td\left( \dfrac{{{t}^{2}}+1}{2} \right)}=\int{t.tdt}=\dfrac{{{t}^{3}}}{3}+C\Rightarrow F\left( x \right)=\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}+C.$
Mà $F\left( 1 \right)=\dfrac{4}{3}\Rightarrow \dfrac{1}{3}+C=\dfrac{4}{3}\Rightarrow C=1\Rightarrow F\left( x \right)=\dfrac{1}{3}\sqrt{{{\left( 2x-1 \right)}^{2}}}+1.$
Đáp án D.