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Câu hỏi: Khẳng định nào sau đây là đúng
A. $\int{\dfrac{1}{{{\cos }^{2}}x}dx=-\cot x+C}$.
B. $\int{{{a}^{x}}dx={{a}^{x}}.\ln a+C}$.
C. $\int{{{e}^{x}}dx=\dfrac{1}{{{e}^{-x}}}+C}$.
D. $\int{\dfrac{1}{x}dx=-\dfrac{1}{{{x}^{2}}}+C}$.
A. $\int{\dfrac{1}{{{\cos }^{2}}x}dx=-\cot x+C}$.
B. $\int{{{a}^{x}}dx={{a}^{x}}.\ln a+C}$.
C. $\int{{{e}^{x}}dx=\dfrac{1}{{{e}^{-x}}}+C}$.
D. $\int{\dfrac{1}{x}dx=-\dfrac{1}{{{x}^{2}}}+C}$.
Ta có: $\int{\dfrac{1}{{{\cos }^{2}}x}dx=\tan x+C}$.
$\int{{{a}^{x}}dx=\dfrac{{{a}^{x}}}{\ln a}+C}$.
$\int{{{e}^{x}}dx={{e}^{x}}+C=\dfrac{1}{{{e}^{-x}}}+C}$.
$\int{\dfrac{1}{x}dx=\ln \left| x \right|+C}$.
$\int{{{a}^{x}}dx=\dfrac{{{a}^{x}}}{\ln a}+C}$.
$\int{{{e}^{x}}dx={{e}^{x}}+C=\dfrac{1}{{{e}^{-x}}}+C}$.
$\int{\dfrac{1}{x}dx=\ln \left| x \right|+C}$.
Đáp án C.