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Câu hỏi: Xét nguyên hàm $\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{x}}+1}}\text{d}x}$ , nếu đặt $t=\sqrt{{{e}^{x}}+1}$ thì $\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{x}}+1}}\text{d}x}$ bằng
A. $\int{2\text{d}t}$.
B. $\int{2{{t}^{2}}\text{d}t}$.
C. $\int{{{t}^{2}}\text{d}t}$.
D. $\int{\dfrac{\text{d}t}{2}}$.
Đặt $t=\sqrt{{{e}^{x}}+1}\Rightarrow {{t}^{2}}={{e}^{x}}+1$
$\Rightarrow 2t\text{d}t={{e}^{x}}\text{d}x$
Khi đó: $\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{x}}+1}}\text{d}x}=\int{\dfrac{2t}{t}\text{d}t}=\int{2\text{d}t}$.
A. $\int{2\text{d}t}$.
B. $\int{2{{t}^{2}}\text{d}t}$.
C. $\int{{{t}^{2}}\text{d}t}$.
D. $\int{\dfrac{\text{d}t}{2}}$.
Đặt $t=\sqrt{{{e}^{x}}+1}\Rightarrow {{t}^{2}}={{e}^{x}}+1$
$\Rightarrow 2t\text{d}t={{e}^{x}}\text{d}x$
Khi đó: $\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{x}}+1}}\text{d}x}=\int{\dfrac{2t}{t}\text{d}t}=\int{2\text{d}t}$.
Đáp án A.