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