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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 0 \right)=0,{f}'\left( x \right)=\dfrac{x}{{{x}^{2}}+1}$. Họ nguyên hàm của hàm số $g\left( x \right)=4x.f\left( x \right)$ là
A. $\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}} \right)-{{x}^{2}}+C$.
B. ${{x}^{2}}\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}$.
C. $\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}+C$.
D. $\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}$.
A. $\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}} \right)-{{x}^{2}}+C$.
B. ${{x}^{2}}\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}$.
C. $\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}+C$.
D. $\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}$.
Ta có $f\left( x \right)=\int{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)+{C}'$
Vì $f\left( 0 \right)=0$ nên ${C}'=0\Rightarrow f\left( x \right)=\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)$
$\Rightarrow \int{g\left( x \right)dx}=\int{2x.\ln \left( {{x}^{2}}+1 \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=\ln \left( {{x}^{2}}+1 \right) \\
& dv=2x.dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2x}{{{x}^{2}}+1}dx \\
& v={{x}^{2}}+1 \\
\end{aligned} \right.$
Nên $\int{g\left( x \right)dx}=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-\int{2xdx}=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}+C$.
Vì $f\left( 0 \right)=0$ nên ${C}'=0\Rightarrow f\left( x \right)=\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)$
$\Rightarrow \int{g\left( x \right)dx}=\int{2x.\ln \left( {{x}^{2}}+1 \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=\ln \left( {{x}^{2}}+1 \right) \\
& dv=2x.dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2x}{{{x}^{2}}+1}dx \\
& v={{x}^{2}}+1 \\
\end{aligned} \right.$
Nên $\int{g\left( x \right)dx}=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-\int{2xdx}=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}+C$.
Đáp án C.