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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$. Gọi $F\left( x \right), G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 2 \right)+G\left( 2 \right)=-4$ và $F\left( 1 \right)+G\left( 1 \right)=9$. Khi đó $\int\limits_{0}^{1}{xf\left( {{x}^{2}}+1 \right)}dx$ bằng
A. $\dfrac{5}{4}$.
B. $13$.
C. $\dfrac{13}{4}$.
D. $\dfrac{-13}{4}$.
A. $\dfrac{5}{4}$.
B. $13$.
C. $\dfrac{13}{4}$.
D. $\dfrac{-13}{4}$.
Đặt $t={{x}^{2}}+1\Rightarrow dt=2xdx$.
Khi đó: $\int\limits_{0}^{1}{xf\left( {{x}^{2}}+1 \right)}dx=\dfrac{1}{2}\int\limits_{1}^{2}{f\left( t \right)}dt=\dfrac{1}{2}\int\limits_{1}^{2}{f\left( x \right)}dx=\dfrac{1}{4}\left[ \int\limits_{1}^{2}{f\left( x \right)}dx+\int\limits_{1}^{2}{f\left( x \right)}dx \right]$
$=\dfrac{1}{4}\left[ F\left( 2 \right)-F\left( 1 \right)+G\left( 2 \right)-G\left( 1 \right) \right]=\dfrac{1}{4}\left( -4-9 \right)=\dfrac{-13}{4}.$
Khi đó: $\int\limits_{0}^{1}{xf\left( {{x}^{2}}+1 \right)}dx=\dfrac{1}{2}\int\limits_{1}^{2}{f\left( t \right)}dt=\dfrac{1}{2}\int\limits_{1}^{2}{f\left( x \right)}dx=\dfrac{1}{4}\left[ \int\limits_{1}^{2}{f\left( x \right)}dx+\int\limits_{1}^{2}{f\left( x \right)}dx \right]$
$=\dfrac{1}{4}\left[ F\left( 2 \right)-F\left( 1 \right)+G\left( 2 \right)-G\left( 1 \right) \right]=\dfrac{1}{4}\left( -4-9 \right)=\dfrac{-13}{4}.$
Đáp án D.