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Câu hỏi: Cho hàm số $f\left( x \right)={{x}^{2}}\sqrt{2{{x}^{3}}+1}$. Một nguyên hàm của hàm số $x{f}'\left( x \right)$ là
A. $\dfrac{1}{9}\left( 7{{x}^{3}}+1 \right)\sqrt{2{{x}^{3}}+1}$.
B. $\dfrac{1}{9}\left( 11{{x}^{3}}+1 \right)\sqrt{2{{x}^{3}}+1}$.
C. $\dfrac{1}{9}\left( 7{{x}^{3}}-1 \right)\sqrt{2{{x}^{3}}+1}$.
D. $\dfrac{1}{9}\left( 11{{x}^{3}}-1 \right)\sqrt{2{{x}^{3}}+1}$.
Ta có $\int{x{f}'\left( x \right)dx}=\int{xd\left( f\left( x \right) \right)=xf\left( x \right)-\int{f\left( x \right)dx}}={{x}^{3}}\sqrt{2{{x}^{3}}+1}-\int{{{x}^{2}}\sqrt{2{{x}^{3}}+1}dx}$ $={{x}^{3}}\sqrt{2{{x}^{3}}+1}-\dfrac{1}{6}\int{\sqrt{2{{x}^{3}}+1}d\left( 2{{x}^{3}}+1 \right)}={{x}^{3}}\sqrt{2{{x}^{3}}+1}-\dfrac{1}{6}.\dfrac{2}{3}\sqrt{{{\left( 2{{x}^{3}}+1 \right)}^{3}}}+C$
$=\dfrac{1}{9}\left( 7{{x}^{3}}-1 \right)\sqrt{2{{x}^{3}}+1}+C$.
A. $\dfrac{1}{9}\left( 7{{x}^{3}}+1 \right)\sqrt{2{{x}^{3}}+1}$.
B. $\dfrac{1}{9}\left( 11{{x}^{3}}+1 \right)\sqrt{2{{x}^{3}}+1}$.
C. $\dfrac{1}{9}\left( 7{{x}^{3}}-1 \right)\sqrt{2{{x}^{3}}+1}$.
D. $\dfrac{1}{9}\left( 11{{x}^{3}}-1 \right)\sqrt{2{{x}^{3}}+1}$.
Ta có $\int{x{f}'\left( x \right)dx}=\int{xd\left( f\left( x \right) \right)=xf\left( x \right)-\int{f\left( x \right)dx}}={{x}^{3}}\sqrt{2{{x}^{3}}+1}-\int{{{x}^{2}}\sqrt{2{{x}^{3}}+1}dx}$ $={{x}^{3}}\sqrt{2{{x}^{3}}+1}-\dfrac{1}{6}\int{\sqrt{2{{x}^{3}}+1}d\left( 2{{x}^{3}}+1 \right)}={{x}^{3}}\sqrt{2{{x}^{3}}+1}-\dfrac{1}{6}.\dfrac{2}{3}\sqrt{{{\left( 2{{x}^{3}}+1 \right)}^{3}}}+C$
$=\dfrac{1}{9}\left( 7{{x}^{3}}-1 \right)\sqrt{2{{x}^{3}}+1}+C$.
Đáp án C.