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Câu hỏi: Cho hàm số $y=f\left( x \right)$ thỏa mãn $f\left( 0 \right)=0;\text{ {f}'}\left( x \right)=\dfrac{x}{{{x}^{2}}+1}$. Họ nguyên hàm của hàm số $g\left( x \right)=4\text{x}f\left( x \right)$ 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: ${f}'\left( x \right)=\dfrac{x}{{{x}^{2}}+1}\Rightarrow f\left( x \right)=\int{{f}'\left( x \right)d\text{x}}=\int{\dfrac{x\text{dx}}{{{x}^{2}}+1}}=\dfrac{1}{2}\int{\dfrac{d\left( {{x}^{2}}+1 \right)}{{{x}^{2}}+1}}=\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)+C$
$f\left( 0 \right)=0\Leftrightarrow \dfrac{1}{2}\ln 1+C=0\Leftrightarrow C=0\Rightarrow f\left( x \right)=\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)$
$\Rightarrow g\left( x \right)=4\text{x}f\left( x \right)=2\text{x}\ln \left( {{x}^{2}}+1 \right)\Rightarrow \int{g\left( x \right)d\text{x}}=\int{2\text{x}\ln \left( {{x}^{2}}+1 \right)d\text{x}}$
Đặt $t={{x}^{2}}+1\Rightarrow dt=2\text{xdx}$
$\Rightarrow \int{g\left( x \right)d\text{x}}=\int{\ln t\text{d}t}=t\ln t-\int{t.\dfrac{1}{t}dt}=t\ln t-\int{dt}=t\ln t-t+C$
$=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-\left( {{x}^{2}}+1 \right)+C$
Đặt $-1+C=c\Rightarrow \int{g\left( x \right)d\text{x}}=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}+c$.
$f\left( 0 \right)=0\Leftrightarrow \dfrac{1}{2}\ln 1+C=0\Leftrightarrow C=0\Rightarrow f\left( x \right)=\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)$
$\Rightarrow g\left( x \right)=4\text{x}f\left( x \right)=2\text{x}\ln \left( {{x}^{2}}+1 \right)\Rightarrow \int{g\left( x \right)d\text{x}}=\int{2\text{x}\ln \left( {{x}^{2}}+1 \right)d\text{x}}$
Đặt $t={{x}^{2}}+1\Rightarrow dt=2\text{xdx}$
$\Rightarrow \int{g\left( x \right)d\text{x}}=\int{\ln t\text{d}t}=t\ln t-\int{t.\dfrac{1}{t}dt}=t\ln t-\int{dt}=t\ln t-t+C$
$=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-\left( {{x}^{2}}+1 \right)+C$
Đặt $-1+C=c\Rightarrow \int{g\left( x \right)d\text{x}}=\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)-{{x}^{2}}+c$.
Đáp án C.