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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 3 \right)=-\dfrac{25}{3}$ và ${f}'\left( x \right)=\dfrac{x}{\sqrt{x+1}-1}$. Khi đó $\int\limits_{3}^{8}{f\left( x \right)dx}$ bằng
A. $\dfrac{68}{5}$.
B. $\dfrac{25}{3}$.
C. $\dfrac{13}{30}$.
D. 10.
A. $\dfrac{68}{5}$.
B. $\dfrac{25}{3}$.
C. $\dfrac{13}{30}$.
D. 10.
Ta có: ${f}'\left( x \right)=\dfrac{x}{\sqrt{x+1}-1}=\dfrac{x\left( \sqrt{x+1}+1 \right)}{\left( \sqrt{x+1}-1 \right)\left( \sqrt{x+1}+1 \right)}=\sqrt{x+1}+1$
$\Rightarrow f\left( x \right)=\int{\left( \sqrt{x+1}+1 \right)dx}=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+x+C$
Do $f\left( 3 \right)=-\dfrac{25}{3}\Rightarrow \dfrac{2}{3}\left( \sqrt{{{\left( 3+1 \right)}^{3}}} \right)+3+C=-\dfrac{25}{3}\Leftrightarrow C=-\dfrac{50}{3}$.
Từ đó: $f\left( x \right)=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+x-\dfrac{50}{3}$
$\Rightarrow \int\limits_{8}^{8}{f\left( x \right)dx}=\int\limits_{3}^{8}{\left[ \dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+x-\dfrac{50}{3} \right]dx}=\left. \left( \dfrac{4}{15}\sqrt[3]{{{\left( x+1 \right)}^{5}}}+\dfrac{{{x}^{2}}}{2}-\dfrac{50}{3} \right) \right|_{3}^{8}=\dfrac{13}{30}$.
Vậy $\int\limits_{3}^{8}{f\left( x \right)dx}=\dfrac{13}{30}$.
$\Rightarrow f\left( x \right)=\int{\left( \sqrt{x+1}+1 \right)dx}=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+x+C$
Do $f\left( 3 \right)=-\dfrac{25}{3}\Rightarrow \dfrac{2}{3}\left( \sqrt{{{\left( 3+1 \right)}^{3}}} \right)+3+C=-\dfrac{25}{3}\Leftrightarrow C=-\dfrac{50}{3}$.
Từ đó: $f\left( x \right)=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+x-\dfrac{50}{3}$
$\Rightarrow \int\limits_{8}^{8}{f\left( x \right)dx}=\int\limits_{3}^{8}{\left[ \dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+x-\dfrac{50}{3} \right]dx}=\left. \left( \dfrac{4}{15}\sqrt[3]{{{\left( x+1 \right)}^{5}}}+\dfrac{{{x}^{2}}}{2}-\dfrac{50}{3} \right) \right|_{3}^{8}=\dfrac{13}{30}$.
Vậy $\int\limits_{3}^{8}{f\left( x \right)dx}=\dfrac{13}{30}$.
Đáp án C.