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Câu hỏi: Tìm $\int{{{x}^{2}}{{\left( 2{{x}^{3}}-1 \right)}^{3}}\text{d}x}$
A. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
B. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
C. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
D. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
A. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
B. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
C. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
D. $\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$.
Đặt $t=2{{x}^{3}}-1\Rightarrow dt=6{{x}^{2}}dx\Rightarrow {{x}^{2}}dx=\dfrac{dt}{6}$
Ta có $\int{{{x}^{2}}{{\left( 2{{x}^{3}}-1 \right)}^{3}}dx}=\dfrac{1}{6}\int{{{t}^{3}}dt}=\dfrac{{{t}^{4}}}{24}+C=\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$
Ta có $\int{{{x}^{2}}{{\left( 2{{x}^{3}}-1 \right)}^{3}}dx}=\dfrac{1}{6}\int{{{t}^{3}}dt}=\dfrac{{{t}^{4}}}{24}+C=\dfrac{{{\left( 2{{x}^{3}}-1 \right)}^{4}}}{24}+C$
Đáp án B.