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Câu hỏi: Kết quả tính $\int{2x\ln (x-1)dx}$ 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. $\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C.$
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. $\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{{{x}^{2}}}{2}-x+C.$
Sử dụng công thức từng phần: $\int\limits_{a}^{b}{udv}=uv\left| \underset{\overset{{}}{\mathop{a}} }{\overset{\overset{{}}{\mathop{b}} }{\mathop{{}}}} -\int\limits_{a}^{b}{vdu} \right.$
$\int{2x\ln \left( x-1 \right)dx}=\int{\ln \left( x-1 \right)d\left( {{x}^{2}} \right)}={{x}^{2}}\ln \left( x-1 \right)-\int{{{x}^{2}}d\left( \ln \left( x-1 \right) \right)}$
$={{x}^{2}}\ln \left( x-1 \right)-\int{{{x}^{2}}.\dfrac{1}{x-1}dx}={{x}^{2}}\ln \left( x-1 \right)-\int{\left( x+1+\dfrac{1}{x-1} \right)dx}$
$={{x}^{2}}\ln \left( x-1 \right)-\dfrac{1}{2}{{x}^{2}}-x-\ln \left| x-1 \right|+C=\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{1}{2}{{x}^{2}}-x+C$
$\int{2x\ln \left( x-1 \right)dx}=\int{\ln \left( x-1 \right)d\left( {{x}^{2}} \right)}={{x}^{2}}\ln \left( x-1 \right)-\int{{{x}^{2}}d\left( \ln \left( x-1 \right) \right)}$
$={{x}^{2}}\ln \left( x-1 \right)-\int{{{x}^{2}}.\dfrac{1}{x-1}dx}={{x}^{2}}\ln \left( x-1 \right)-\int{\left( x+1+\dfrac{1}{x-1} \right)dx}$
$={{x}^{2}}\ln \left( x-1 \right)-\dfrac{1}{2}{{x}^{2}}-x-\ln \left| x-1 \right|+C=\left( {{x}^{2}}-1 \right)\ln \left( x-1 \right)-\dfrac{1}{2}{{x}^{2}}-x+C$
Đáp án D.