Cho nguyên hàm (x-1)/(x^3-3x-4)dx=aln(x-4)+bln(x+1)+c trên khoảng

Đáp án A

Đồng nhất thức \(\frac{{x - 1}}{{{x^2} - 3{\rm{x}} - 4}} = \frac{{x - 1}}{{\left( {x - 4} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 1}}\).

Suy ra$x-1=A\left(x+1\right)+B\left(x-4\right)\iff \left\{\begin{array}{l}A+B=1\\ A-4B=-1\end{array}\right.\iff \left\{\begin{array}{l}A=\frac{3}{5}\\ B=\frac{2}{5}\end{array}\right.$

Do đó $\int \frac{x-1}{x^2-3\text{x}-4}d\text{x}=\int \left(\frac{3}{5}.\frac{1}{x-4}+\frac{2}{5}.\frac{1}{x+1}\right)d\text{x}=\frac{3}{5}\ln \left(x-4\right)+\frac{2}{5}\ln \left(x+1\right)+C$

Suy ra $a=\frac{3}{5},b=\frac{2}{5}\Rightarrow T=\frac{13}{5}$