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Câu hỏi: Họ nguyên hàm của hàm số $f(x)={{2017}^{x}}+2018\text{x}-2019$ là
A. $\int{f(x)d\text{x}}=\dfrac{{{2017}^{x}}}{\ln 2017}+1009{{\text{x}}^{2}}-2019\text{x}+C$
B. $\int{f(x)d\text{x}}=\dfrac{{{2017}^{x}}}{\ln 2017}+2018{{x}^{2}}-2019\text{x}+C$
C. $\int{f(x)d\text{x}}={{2017}^{x}}.\ln 2+1009{{\text{x}}^{2}}-2019\text{x}+C$
D. $\int{f(x)d\text{x}}={{2017}^{x}}.\ln 2+1009{{\text{x}}^{2}}-2019\text{x}+C$
A. $\int{f(x)d\text{x}}=\dfrac{{{2017}^{x}}}{\ln 2017}+1009{{\text{x}}^{2}}-2019\text{x}+C$
B. $\int{f(x)d\text{x}}=\dfrac{{{2017}^{x}}}{\ln 2017}+2018{{x}^{2}}-2019\text{x}+C$
C. $\int{f(x)d\text{x}}={{2017}^{x}}.\ln 2+1009{{\text{x}}^{2}}-2019\text{x}+C$
D. $\int{f(x)d\text{x}}={{2017}^{x}}.\ln 2+1009{{\text{x}}^{2}}-2019\text{x}+C$
Ta có $\int{f(x)d\text{x}}=\int{\left( {{2017}^{x}}+2018x-2019 \right)d\text{x}}=\int{{{2017}^{x}}d\text{x}}+2018\int{x\text{dx}}-2019\int{d\text{x}}$
$=\dfrac{{{2017}^{x}}}{\ln 2017}+1009{{\text{x}}^{2}}-2019\text{x}+C$.
$=\dfrac{{{2017}^{x}}}{\ln 2017}+1009{{\text{x}}^{2}}-2019\text{x}+C$.
Đáp án A.