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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và thỏa $\int\limits_{-2}^{2}{f\left( \sqrt{{{x}^{2}}+5}-x \right)\text{d}x}=1,$
$\int\limits_{1}^{5}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x}=3.$ Tính $\int\limits_{1}^{5}{f\left( x \right)\text{d}x}.$
A. 0.
B. -15.
C. -2.
D. -13.
$\int\limits_{1}^{5}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x}=3.$ Tính $\int\limits_{1}^{5}{f\left( x \right)\text{d}x}.$
A. 0.
B. -15.
C. -2.
D. -13.
Đặt: $t=\sqrt{{{x}^{2}}+5}-x\Rightarrow x=\dfrac{5-{{t}^{2}}}{2t}\Rightarrow \text{d}x=-\left( \dfrac{1}{2}+\dfrac{5}{2{{t}^{2}}} \right)\text{d}t$.
Ta có: $1=\int\limits_{1}^{5}{f\left( t \right)}\left( \dfrac{1}{2}+\dfrac{5}{2{{t}^{2}}} \right)\text{d}t=\dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)}\text{d}t+\dfrac{5}{2}\int\limits_{1}^{5}{\dfrac{f\left( t \right)}{{{t}^{2}}}\text{d}t}$
$\Rightarrow \dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)}\text{d}t=1-\dfrac{5}{2}\int\limits_{1}^{5}{\dfrac{f\left( t \right)}{{{t}^{2}}}\text{d}t}=1-\dfrac{5}{2}.3=-\dfrac{13}{2}$
$\Rightarrow \int\limits_{1}^{5}{f\left( t \right)}\text{d}t=-13$
Ta có: $1=\int\limits_{1}^{5}{f\left( t \right)}\left( \dfrac{1}{2}+\dfrac{5}{2{{t}^{2}}} \right)\text{d}t=\dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)}\text{d}t+\dfrac{5}{2}\int\limits_{1}^{5}{\dfrac{f\left( t \right)}{{{t}^{2}}}\text{d}t}$
$\Rightarrow \dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)}\text{d}t=1-\dfrac{5}{2}\int\limits_{1}^{5}{\dfrac{f\left( t \right)}{{{t}^{2}}}\text{d}t}=1-\dfrac{5}{2}.3=-\dfrac{13}{2}$
$\Rightarrow \int\limits_{1}^{5}{f\left( t \right)}\text{d}t=-13$
Đáp án D.