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Câu hỏi: Tìm nguyên hàm của hàm số $f\left( x \right)={{e}^{x}}\left( 1-\dfrac{2{{e}^{-x}}}{{{x}^{5}}} \right)$.
A. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}+\dfrac{1}{2{{x}^{4}}}+C$
B. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}-\dfrac{1}{2{{x}^{4}}}+C$
C. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}-\dfrac{2}{{{x}^{4}}}+C$
D. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}+\dfrac{2}{{{x}^{4}}}+C$
A. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}+\dfrac{1}{2{{x}^{4}}}+C$
B. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}-\dfrac{1}{2{{x}^{4}}}+C$
C. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}-\dfrac{2}{{{x}^{4}}}+C$
D. $\int{f\left( x \right)}\text{ d}x={{e}^{x}}+\dfrac{2}{{{x}^{4}}}+C$
$\int{f\left( x \right)}\text{ d}x=\int{{{e}^{x}}\left( 1-\dfrac{2{{e}^{-x}}}{{{x}^{5}}} \right)\text{d}x=\int{\left( {{e}^{x}}-\dfrac{2}{{{x}^{5}}} \right)}}\text{ d}x={{e}^{x}}+\dfrac{1}{2{{x}^{4}}}+C$.
Đáp án A.