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Câu hỏi: Kết quả tính $\int{2x\ln \left( x-1 \right)\text{d}x}$ bằng:
A. $\left( {{x}^{2}}+1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
B. $\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}+x+C$.
C. ${{x}^{2}}\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
D. $\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
Xét $\int{2x\ln \left( x-1 \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( x-1 \right) \\
& \text{d}v=2x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x-1}\text{d}x \\
& v={{x}^{2}}-1 \\
\end{aligned} \right.$.
$\Rightarrow \int{2x\ln \left( x-1 \right)\text{d}x}=\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\int{\left( x+1 \right)\text{d}x}=\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
A. $\left( {{x}^{2}}+1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
B. $\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}+x+C$.
C. ${{x}^{2}}\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
D. $\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
Xét $\int{2x\ln \left( x-1 \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( x-1 \right) \\
& \text{d}v=2x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x-1}\text{d}x \\
& v={{x}^{2}}-1 \\
\end{aligned} \right.$.
$\Rightarrow \int{2x\ln \left( x-1 \right)\text{d}x}=\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\int{\left( x+1 \right)\text{d}x}=\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C$.
Đáp án D.