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Câu hỏi: Cho $\int{f\left( x \right)dx}=\sqrt{{{x}^{2}}+4}.{{e}^{2x-1}}+C$. Tìm $\int{f\left( 2x \right)dx}$.
A. $\int{f\left( 2x \right)dx}=2\sqrt{{{x}^{2}}+1}.{{e}^{4x-1}}+C$.
B. $\int{f\left( 2x \right)dx}=\dfrac{1}{2}\sqrt{{{x}^{2}}+16}.{{e}^{x-1}}+C$.
C. $\int{f\left( 2x \right)dx}=\sqrt{{{x}^{2}}+4}.{{e}^{4x-1}}+C$.
D. $\int{f\left( 2x \right)dx}=\sqrt{{{x}^{2}}+1}.{{e}^{4x-1}}+C$.
A. $\int{f\left( 2x \right)dx}=2\sqrt{{{x}^{2}}+1}.{{e}^{4x-1}}+C$.
B. $\int{f\left( 2x \right)dx}=\dfrac{1}{2}\sqrt{{{x}^{2}}+16}.{{e}^{x-1}}+C$.
C. $\int{f\left( 2x \right)dx}=\sqrt{{{x}^{2}}+4}.{{e}^{4x-1}}+C$.
D. $\int{f\left( 2x \right)dx}=\sqrt{{{x}^{2}}+1}.{{e}^{4x-1}}+C$.
$\int{f\left( x \right)dx}=\sqrt{{{x}^{2}}+4}.{{e}^{2x-1}}+C$
Đặt $x=2t$, ta có $\int{f\left( 2t \right)d\left( 2t \right)}=\sqrt{{{\left( 2t \right)}^{2}}+4}{{e}^{2\left( 2t \right)-1}}+C=\sqrt{4{{t}^{2}}+4}.{{e}^{4t-1}}+C$
$\Leftrightarrow \int{f\left( 2t \right)dt}=\sqrt{{{t}^{2}}+1}.{{e}^{4t-1}}+\dfrac{1}{2}C$
Vậy $\int{f\left( 2x \right)dx}=\sqrt{{{x}^{2}}+1}.{{e}^{4x-1}}+C$.
Đặt $x=2t$, ta có $\int{f\left( 2t \right)d\left( 2t \right)}=\sqrt{{{\left( 2t \right)}^{2}}+4}{{e}^{2\left( 2t \right)-1}}+C=\sqrt{4{{t}^{2}}+4}.{{e}^{4t-1}}+C$
$\Leftrightarrow \int{f\left( 2t \right)dt}=\sqrt{{{t}^{2}}+1}.{{e}^{4t-1}}+\dfrac{1}{2}C$
Vậy $\int{f\left( 2x \right)dx}=\sqrt{{{x}^{2}}+1}.{{e}^{4x-1}}+C$.
Đáp án D.