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Câu hỏi: Tìm một nguyên hàm $F\left( x \right)$ của hàm số
$f\left( x \right)=ax+\dfrac{b}{{{x}^{2}}} \left( x\ne 0 \right)$ biết $F\left( -1 \right)= 3, F\left( 1 \right)=5, f\left( 1 \right)=-3.$
A. $F\left( x \right)=\dfrac{3{{x}^{2}}}{2}-\dfrac{3}{2x}$.
B. $F\left( x \right)=\dfrac{{{x}^{2}}}{2}+\dfrac{3}{x}+\dfrac{3}{2}$.
C. $F\left( x \right)=-{{x}^{2}}+\dfrac{1}{x}+5$.
D. $F\left( x \right)=-2{{x}^{2}}+\dfrac{1}{2x}+5$.
$f\left( x \right)=ax+\dfrac{b}{{{x}^{2}}} \left( x\ne 0 \right)$ biết $F\left( -1 \right)= 3, F\left( 1 \right)=5, f\left( 1 \right)=-3.$
A. $F\left( x \right)=\dfrac{3{{x}^{2}}}{2}-\dfrac{3}{2x}$.
B. $F\left( x \right)=\dfrac{{{x}^{2}}}{2}+\dfrac{3}{x}+\dfrac{3}{2}$.
C. $F\left( x \right)=-{{x}^{2}}+\dfrac{1}{x}+5$.
D. $F\left( x \right)=-2{{x}^{2}}+\dfrac{1}{2x}+5$.
$F\left( x \right)=\int{f\left( x \right)}\text{d}x=\int{\left( ax+\dfrac{b}{{{x}^{2}}} \right)}\text{d}x=\dfrac{a{{x}^{2}}}{2}-\dfrac{b}{x}+C.$
Theo giả thiết ta có: $F\left( -1 \right)= 3\Leftrightarrow \dfrac{1}{2}a+b+C=3\left( 1 \right)$
$F\left( 1 \right)=5\Leftrightarrow \dfrac{1}{2}a-b+C=5\left( 2 \right)$
$f\left( 1 \right)=-3\Leftrightarrow a+b=-3 \left( 3 \right)$
Từ (1), (2) và (3) suy ra $a=-2, b=-1, c=5.$ Vậy $F\left( x \right)=-{{x}^{2}}+\dfrac{1}{x}+5$.
Theo giả thiết ta có: $F\left( -1 \right)= 3\Leftrightarrow \dfrac{1}{2}a+b+C=3\left( 1 \right)$
$F\left( 1 \right)=5\Leftrightarrow \dfrac{1}{2}a-b+C=5\left( 2 \right)$
$f\left( 1 \right)=-3\Leftrightarrow a+b=-3 \left( 3 \right)$
Từ (1), (2) và (3) suy ra $a=-2, b=-1, c=5.$ Vậy $F\left( x \right)=-{{x}^{2}}+\dfrac{1}{x}+5$.
Đáp án C.