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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=\sin x+2{{e}^{x}}-5x$ là
A. $\mathop{\int }^{}f\left( x \right)dx=\cos x+2{{e}^{x}}-5+C.$
B. $\mathop{\int }^{}f\left( x \right)dx=-\cos x+2{{e}^{x+1}}-\dfrac{5{{x}^{2}}}{2}+C.$
C. $\mathop{\int }^{}f\left( x \right)dx=-\cos x+2{{e}^{x}}-\dfrac{5{{x}^{2}}}{2}+C.$
D. $\mathop{\int }^{}f\left( x \right)dx=\cos x+2{{e}^{x}}-\dfrac{5{{x}^{2}}}{2}+C.$
A. $\mathop{\int }^{}f\left( x \right)dx=\cos x+2{{e}^{x}}-5+C.$
B. $\mathop{\int }^{}f\left( x \right)dx=-\cos x+2{{e}^{x+1}}-\dfrac{5{{x}^{2}}}{2}+C.$
C. $\mathop{\int }^{}f\left( x \right)dx=-\cos x+2{{e}^{x}}-\dfrac{5{{x}^{2}}}{2}+C.$
D. $\mathop{\int }^{}f\left( x \right)dx=\cos x+2{{e}^{x}}-\dfrac{5{{x}^{2}}}{2}+C.$
Ta có $\mathop{\int }^{}f\left( x \right)dx=\mathop{\int }^{}\left( \sin x+2{{e}^{x}}-5x \right)dx=-\cos x+2{{e}^{x}}-\dfrac{5{{x}^{2}}}{2}+C.$
Đáp án C.