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Câu hỏi: Họ các nguyên hàm của hàm số $f\left( x \right)=\dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}$ trên khoảng $\left( 0;+\infty \right)$ là
A. $\dfrac{1}{x+1}\left( x+\ln x \right)-\ln x+C$.
B. $-\dfrac{1}{x+1}\left( x+\ln x \right)+\ln x+C$.
C. $\dfrac{1}{x+1}\left( x+\ln x \right)+\ln x+C$.
D. $-\dfrac{1}{x+1}\left( x+\ln x \right)-\ln x+C$.
A. $\dfrac{1}{x+1}\left( x+\ln x \right)-\ln x+C$.
B. $-\dfrac{1}{x+1}\left( x+\ln x \right)+\ln x+C$.
C. $\dfrac{1}{x+1}\left( x+\ln x \right)+\ln x+C$.
D. $-\dfrac{1}{x+1}\left( x+\ln x \right)-\ln x+C$.
Đặt $\left\{ \begin{aligned}
& u=x+\ln x \\
& dv=\dfrac{dx}{{{\left( x+1 \right)}^{2}}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( 1+\dfrac{1}{x} \right)dx \\
& v=-\dfrac{1}{x+1} \\
\end{aligned} \right.$.
$\begin{aligned}
& \Rightarrow I=\int{\dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}dx}=-\dfrac{1}{x+1}\left( x+\ln x \right)+\int{\dfrac{1}{x+1}.\dfrac{x+1}{x}dx}=-\dfrac{1}{x+1}\left( x+\ln x \right)+\int{\dfrac{dx}{x}} \\
& =-\dfrac{1}{x+1}\left( x+\ln x \right)+\ln x+C \\
\end{aligned}$
& u=x+\ln x \\
& dv=\dfrac{dx}{{{\left( x+1 \right)}^{2}}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( 1+\dfrac{1}{x} \right)dx \\
& v=-\dfrac{1}{x+1} \\
\end{aligned} \right.$.
$\begin{aligned}
& \Rightarrow I=\int{\dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}dx}=-\dfrac{1}{x+1}\left( x+\ln x \right)+\int{\dfrac{1}{x+1}.\dfrac{x+1}{x}dx}=-\dfrac{1}{x+1}\left( x+\ln x \right)+\int{\dfrac{dx}{x}} \\
& =-\dfrac{1}{x+1}\left( x+\ln x \right)+\ln x+C \\
\end{aligned}$
Đáp án B.