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Câu hỏi: Cho nguyên hàm $\int{\dfrac{x-1}{{{x}^{2}}-3x-4}dx}=a\ln \left( x-4 \right)+b\ln \left( x+1 \right)+C$ trên khoảng $\left( 4;+\infty \right)$. Tính giá trị của biểu thức $T=3a+2b$.
A. $T=\dfrac{13}{5}$.
B. $T=\dfrac{12}{5}$.
C. $T=0$.
D. $T=1$.
A. $T=\dfrac{13}{5}$.
B. $T=\dfrac{12}{5}$.
C. $T=0$.
D. $T=1$.
Đồng nhất thức $\dfrac{x-1}{{{x}^{2}}-3x-4}=\dfrac{x-1}{\left( x-4 \right)\left( x+1 \right)}=\dfrac{A}{x-4}+\dfrac{B}{x+1}$
Suy ra $x-1=A\left( x+1 \right)+B\left( x-4 \right)\Leftrightarrow \left\{ \begin{aligned}
& A+B=1 \\
& A-4B=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=\dfrac{3}{5} \\
& B=\dfrac{2}{5} \\
\end{aligned} \right.$
Do đó $\int{\dfrac{x-1}{{{x}^{2}}-3x-4}dx}=\int{\left( \dfrac{3}{5}.\dfrac{1}{x-4}+\dfrac{2}{5}.\dfrac{1}{x+1} \right)dx}=\dfrac{3}{5}\ln \left( x-4 \right)+\dfrac{2}{5}\ln \left( x+1 \right)+C$
Suy ra $a=\dfrac{3}{5},b=\dfrac{2}{5}\Rightarrow T=\dfrac{13}{5}$.
Suy ra $x-1=A\left( x+1 \right)+B\left( x-4 \right)\Leftrightarrow \left\{ \begin{aligned}
& A+B=1 \\
& A-4B=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=\dfrac{3}{5} \\
& B=\dfrac{2}{5} \\
\end{aligned} \right.$
Do đó $\int{\dfrac{x-1}{{{x}^{2}}-3x-4}dx}=\int{\left( \dfrac{3}{5}.\dfrac{1}{x-4}+\dfrac{2}{5}.\dfrac{1}{x+1} \right)dx}=\dfrac{3}{5}\ln \left( x-4 \right)+\dfrac{2}{5}\ln \left( x+1 \right)+C$
Suy ra $a=\dfrac{3}{5},b=\dfrac{2}{5}\Rightarrow T=\dfrac{13}{5}$.
Đáp án A.