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