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Câu hỏi: . Tính nguyên hàm $I=\int{\dfrac{x-5}{{{x}^{2}}-1}\text{d}x}$
A. $I=\dfrac{3}{2}\ln \left| \dfrac{x+1}{x-1} \right|+C.$
B. $I=\dfrac{3}{2}\ln \left| \dfrac{x-1}{x+1} \right|+C.$
C. $I=\ln \left| \dfrac{{{\left( x+1 \right)}^{3}}}{{{\left( x-1 \right)}^{2}}} \right|+C.$
D. $I=\ln \left| \dfrac{{{\left( x+1 \right)}^{2}}}{{{\left( x-1 \right)}^{3}}} \right|+C.$
A. $I=\dfrac{3}{2}\ln \left| \dfrac{x+1}{x-1} \right|+C.$
B. $I=\dfrac{3}{2}\ln \left| \dfrac{x-1}{x+1} \right|+C.$
C. $I=\ln \left| \dfrac{{{\left( x+1 \right)}^{3}}}{{{\left( x-1 \right)}^{2}}} \right|+C.$
D. $I=\ln \left| \dfrac{{{\left( x+1 \right)}^{2}}}{{{\left( x-1 \right)}^{3}}} \right|+C.$
Ta có: $\dfrac{x-5}{{{x}^{2}}-1}=\dfrac{A}{x-1}+\dfrac{B}{x+1}=\dfrac{\left( A+B \right)x+A-B}{{{x}^{2}}-1}$
Đồng nhất 2 vế ta có: $\left\{ \begin{aligned}
& A+B=1 \\
& A-B=-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=-2 \\
& B=3 \\
\end{aligned} \right.$
Suy ra $I=\int{\left( \dfrac{3}{x+1}-\dfrac{2}{x-1} \right)d\text{x}}=3\ln \left| x+1 \right|-2\ln \left| x-1 \right|+C=\ln \left| \dfrac{{{\left( x+1 \right)}^{3}}}{{{\left( x-1 \right)}^{2}}} \right|+C$.
Đồng nhất 2 vế ta có: $\left\{ \begin{aligned}
& A+B=1 \\
& A-B=-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=-2 \\
& B=3 \\
\end{aligned} \right.$
Suy ra $I=\int{\left( \dfrac{3}{x+1}-\dfrac{2}{x-1} \right)d\text{x}}=3\ln \left| x+1 \right|-2\ln \left| x-1 \right|+C=\ln \left| \dfrac{{{\left( x+1 \right)}^{3}}}{{{\left( x-1 \right)}^{2}}} \right|+C$.
Đáp án C.