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Câu hỏi: Nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=2x+\dfrac{1}{{{\sin }^{2}}x}$ thỏa mãn $F\left( \dfrac{\pi }{4} \right)=-1$ là
A. $-\cot x+{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}$.
B. $\cot x-{{x}^{2}}+\dfrac{{{\pi }^{2}}}{16}$.
C. $-\cot x+{{x}^{2}}-1$.
D. $\cot x+{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}$.
A. $-\cot x+{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}$.
B. $\cot x-{{x}^{2}}+\dfrac{{{\pi }^{2}}}{16}$.
C. $-\cot x+{{x}^{2}}-1$.
D. $\cot x+{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}$.
Ta có $F(x)=\int{\left( 2x+\dfrac{1}{{{\sin }^{2}}x} \right)dx={{x}^{2}}-\cot x+C}$
$F\left( \dfrac{\pi }{4} \right)=-1\Leftrightarrow {{\left( \dfrac{\pi }{4} \right)}^{2}}-\cot \dfrac{\pi }{4}+C=-1\Leftrightarrow C=-\dfrac{{{\pi }^{2}}}{16}$
Vậy $F(x)=-\cot x+{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}$.
$F\left( \dfrac{\pi }{4} \right)=-1\Leftrightarrow {{\left( \dfrac{\pi }{4} \right)}^{2}}-\cot \dfrac{\pi }{4}+C=-1\Leftrightarrow C=-\dfrac{{{\pi }^{2}}}{16}$
Vậy $F(x)=-\cot x+{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}$.
Đáp án A.