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Câu hỏi: Cho hàm số $f\left( x \right)=x\sqrt{{{x}^{2}}+1}$. Họ tất cả các nguyên hàm của hàm số $g\left( x \right)=x.f'\left( x \right)$ là
A. $\dfrac{3}{2}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
B. $\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
C. $\dfrac{2}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
D. $\dfrac{2}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}+1}+C.$
A. $\dfrac{3}{2}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
B. $\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
C. $\dfrac{2}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
D. $\dfrac{2}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}+1}+C.$
Ta có: $\int\limits_{{}}^{{}}{g\left( x \right)dx}=\int\limits_{{}}^{{}}{xf'\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\int\limits_{{}}^{{}}{g\left( x \right)dx}=\int\limits_{{}}^{{}}{xf'\left( x \right)dx}=xf\left( x \right)-\int\limits_{{}}^{{}}{f\left( x \right)dx}={{x}^{2}}\sqrt{{{x}^{2}}+1}-\int\limits_{{}}^{{}}{x\sqrt{{{x}^{2}}+1}dx}$
Tính $I=\int\limits_{{}}^{{}}{x\sqrt{{{x}^{2}}+1}dx}$
$\sqrt{{{x}^{2}}+1}=t\Rightarrow {{x}^{2}}+1={{t}^{2}}\Rightarrow xdx=tdt$
Khi đó: $I=\int\limits_{{}}^{{}}{{{t}^{2}}dt}=\dfrac{{{t}^{3}}}{3}+C=\dfrac{\sqrt{{{\left( {{x}^{2}}+1 \right)}^{3}}}}{3}+C$
$\int\limits_{{}}^{{}}{g\left( x \right)dx}={{x}^{2}}\sqrt{{{x}^{2}}+1}-\dfrac{1}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}+C=\dfrac{2}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\int\limits_{{}}^{{}}{g\left( x \right)dx}=\int\limits_{{}}^{{}}{xf'\left( x \right)dx}=xf\left( x \right)-\int\limits_{{}}^{{}}{f\left( x \right)dx}={{x}^{2}}\sqrt{{{x}^{2}}+1}-\int\limits_{{}}^{{}}{x\sqrt{{{x}^{2}}+1}dx}$
Tính $I=\int\limits_{{}}^{{}}{x\sqrt{{{x}^{2}}+1}dx}$
$\sqrt{{{x}^{2}}+1}=t\Rightarrow {{x}^{2}}+1={{t}^{2}}\Rightarrow xdx=tdt$
Khi đó: $I=\int\limits_{{}}^{{}}{{{t}^{2}}dt}=\dfrac{{{t}^{3}}}{3}+C=\dfrac{\sqrt{{{\left( {{x}^{2}}+1 \right)}^{3}}}}{3}+C$
$\int\limits_{{}}^{{}}{g\left( x \right)dx}={{x}^{2}}\sqrt{{{x}^{2}}+1}-\dfrac{1}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}+C=\dfrac{2}{3}\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}+1}+C.$
Đáp án C.