Biết \(\int\limits^4_3\dfrac{\text{d}x}{x^2+x}=a\ln2+b\ln3+c\ln5\), với \(a,b,c\) là các số nguyên. Tổng \(a+b+c\) bằng
\(6\).\(2\).\(-2\).\(0\).Hướng dẫn giải:\(\int\limits^4_3\dfrac{\text{d}x}{x^2+x}=\int\limits^4_3\dfrac{1}{x\left(x+1\right)}\text{d}x=\int\limits^4_3\left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)\text{d}x\)
\(=\int\limits^4_3\dfrac{\text{d}x}{x}-\int\limits^4_3\dfrac{\text{d}\left(x+1\right)}{x+1}=\left[\ln x-\ln\left(x+1\right)\right]|^4_3\)
\(=\ln4-\ln5-\ln3+\ln4=2\ln4-\ln3-\ln5=4\ln2-\ln3-\ln5\)
Vậy \(a=4,b=-1,c=-1,S=a+b+c=2\).
