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Câu hỏi: Cho hàm số $f\left( x \right)$, biết $f\left( 1 \right)=1,f'\left( x \right)=\dfrac{2x}{3x+1-\sqrt{3x+1}}, x>0$. Khi đó $\int\limits_{1}^{5}{f\left( x \right)\text{d}}x$ bằng
A. $\dfrac{128}{9}$.
B. $\dfrac{184}{9}$.
C. $\dfrac{440}{27}$.
D. $\dfrac{916}{81}$.
A. $\dfrac{128}{9}$.
B. $\dfrac{184}{9}$.
C. $\dfrac{440}{27}$.
D. $\dfrac{916}{81}$.
Ta có $f\left( x \right)=\int{\dfrac{2x}{3x+1-\sqrt{3x+1}}\text{d}x}$
Đặt $t=\sqrt{3x+1}\Rightarrow x=\dfrac{{{t}^{2}}-1}{3}$ $\Rightarrow dx=\dfrac{2}{3}t.\text{d}t$.
$f\left( x \right)=\dfrac{4}{9}\int{\dfrac{\left( {{t}^{2}}-1 \right)t}{{{t}^{2}}-t}}\text{d}t=\dfrac{4}{9}\int{\left( t+1 \right)\text{d}t}$ = $\dfrac{2}{9}{{t}^{2}}+\dfrac{4}{9}t+C=\dfrac{2}{9}\left( 3x+1 \right)+\dfrac{4}{9}\sqrt{3x+1}+C$
Có $f\left( 1 \right)=1\Rightarrow C=-\dfrac{7}{9}$ $\Rightarrow f\left( x \right)=\dfrac{2x}{3}+\dfrac{4}{9}\sqrt{3x+1}-\dfrac{5}{9}$.
Khi đó
$\int\limits_{1}^{5}{f\left( x \right)\text{d}}x=\int\limits_{1}^{5}{\left( \dfrac{2x}{3}+\dfrac{4}{9}\sqrt{3x+1}-\dfrac{5}{9} \right)}\text{d}x=\left. \left[ \dfrac{1}{3}{{x}^{2}}+\dfrac{8}{81}\left( 3x+1 \right)\sqrt{3x+1}-\dfrac{5}{9}x \right] \right|_{1}^{5}$ $=\dfrac{916}{81}$. Không có đáp án. Nên đã sửa đáp ánD.
Đặt $t=\sqrt{3x+1}\Rightarrow x=\dfrac{{{t}^{2}}-1}{3}$ $\Rightarrow dx=\dfrac{2}{3}t.\text{d}t$.
$f\left( x \right)=\dfrac{4}{9}\int{\dfrac{\left( {{t}^{2}}-1 \right)t}{{{t}^{2}}-t}}\text{d}t=\dfrac{4}{9}\int{\left( t+1 \right)\text{d}t}$ = $\dfrac{2}{9}{{t}^{2}}+\dfrac{4}{9}t+C=\dfrac{2}{9}\left( 3x+1 \right)+\dfrac{4}{9}\sqrt{3x+1}+C$
Có $f\left( 1 \right)=1\Rightarrow C=-\dfrac{7}{9}$ $\Rightarrow f\left( x \right)=\dfrac{2x}{3}+\dfrac{4}{9}\sqrt{3x+1}-\dfrac{5}{9}$.
Khi đó
$\int\limits_{1}^{5}{f\left( x \right)\text{d}}x=\int\limits_{1}^{5}{\left( \dfrac{2x}{3}+\dfrac{4}{9}\sqrt{3x+1}-\dfrac{5}{9} \right)}\text{d}x=\left. \left[ \dfrac{1}{3}{{x}^{2}}+\dfrac{8}{81}\left( 3x+1 \right)\sqrt{3x+1}-\dfrac{5}{9}x \right] \right|_{1}^{5}$ $=\dfrac{916}{81}$. Không có đáp án. Nên đã sửa đáp ánD.
Đáp án D.