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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}}$. Họ tất cả các nguyên hàm của hàm số $g\left( x \right)=\left( x+1 \right){f}'\left( x \right)$
A. $\dfrac{x-1}{\sqrt{{{x}^{2}}+1}}+C$
B. $\dfrac{{{x}^{2}}+2x-1}{2\sqrt{{{x}^{2}}+1}}+C$
C. $\dfrac{2{{x}^{2}}+x+1}{\sqrt{{{x}^{2}}+1}}+C$
D. $\dfrac{x+1}{\sqrt{{{x}^{2}}+1}}+C$
A. $\dfrac{x-1}{\sqrt{{{x}^{2}}+1}}+C$
B. $\dfrac{{{x}^{2}}+2x-1}{2\sqrt{{{x}^{2}}+1}}+C$
C. $\dfrac{2{{x}^{2}}+x+1}{\sqrt{{{x}^{2}}+1}}+C$
D. $\dfrac{x+1}{\sqrt{{{x}^{2}}+1}}+C$
Ta có: $\int{g\left( x \right) dx}=\int{\left( x+1 \right){f}'\left( x \right) dx=}\int{\left( x+1 \right) d\left( f\left( x \right) \right)=}\left( x+1 \right).f\left( x \right)-\int{f\left( x \right).dx}$
$=\left( x+1 \right).\dfrac{x}{\sqrt{{{x}^{2}}+1}}-\int{\dfrac{x}{\sqrt{{{x}^{2}}+1}}.dx}$
$=\left( x+1 \right).\dfrac{x}{\sqrt{{{x}^{2}}+1}}-\sqrt{{{x}^{2}}+1}+C$
$=\dfrac{x-1}{\sqrt{{{x}^{2}}+1}}+C$.
$=\left( x+1 \right).\dfrac{x}{\sqrt{{{x}^{2}}+1}}-\int{\dfrac{x}{\sqrt{{{x}^{2}}+1}}.dx}$
$=\left( x+1 \right).\dfrac{x}{\sqrt{{{x}^{2}}+1}}-\sqrt{{{x}^{2}}+1}+C$
$=\dfrac{x-1}{\sqrt{{{x}^{2}}+1}}+C$.
Đáp án A.