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Câu hỏi: Cho $f\left( \dfrac{3x-4}{3x+4} \right)=x+2$. Khi đó $I=\int{f\left( x \right)dx}$ bằng
A. $I={{e}^{x+2}}\ln \left| \dfrac{3x-4}{3x+4} \right|+C$.
B. $I=-\dfrac{8}{3}\ln \left| 1-x \right|+\dfrac{2}{3}x+C$.
C. $I=\dfrac{8}{3}\ln \left| x-1 \right|+\dfrac{x}{3}+C$.
D. $I=\dfrac{8}{3}\ln \left| x-1 \right|+x+C$.
A. $I={{e}^{x+2}}\ln \left| \dfrac{3x-4}{3x+4} \right|+C$.
B. $I=-\dfrac{8}{3}\ln \left| 1-x \right|+\dfrac{2}{3}x+C$.
C. $I=\dfrac{8}{3}\ln \left| x-1 \right|+\dfrac{x}{3}+C$.
D. $I=\dfrac{8}{3}\ln \left| x-1 \right|+x+C$.
Đặt: $\dfrac{3x-4}{3x+4}=t\Leftrightarrow 1-\dfrac{8}{3x+4}=t\Leftrightarrow \dfrac{1}{3x+4}=\dfrac{1-t}{8}\Leftrightarrow x=\dfrac{4}{3}.\dfrac{1+t}{1-t}$
Theo giả thiết: $f\left( t \right)=\dfrac{4}{3}.\dfrac{1+t}{1-t}+2=\dfrac{10-2t}{3\left( 1-t \right)}=\dfrac{2}{3}+\dfrac{8}{3\left( 1-t \right)}$
Nên: $f\left( x \right)=\dfrac{2}{3}+\dfrac{8}{3}.\dfrac{1}{1-x}\Rightarrow \int{f\left( x \right)dx=\dfrac{2}{3}x-\dfrac{8}{3}\ln \left| 1-x \right|+}C$
Theo giả thiết: $f\left( t \right)=\dfrac{4}{3}.\dfrac{1+t}{1-t}+2=\dfrac{10-2t}{3\left( 1-t \right)}=\dfrac{2}{3}+\dfrac{8}{3\left( 1-t \right)}$
Nên: $f\left( x \right)=\dfrac{2}{3}+\dfrac{8}{3}.\dfrac{1}{1-x}\Rightarrow \int{f\left( x \right)dx=\dfrac{2}{3}x-\dfrac{8}{3}\ln \left| 1-x \right|+}C$
Đáp án B.