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