Biết \(\int\limits^1_0\dfrac{\left(3x-1\right)\text{d}x}{x^2+6x+9}=3\ln\dfrac{a}{b}-\dfrac{5}{6}\), trong đó \(a,b\) nguyên dương và \(\frac{a}{b}\) là phân số tối giản
Tích \(ab\) bằng
\(-5\). \(12\). \(6\). \(\frac{5}{4}\). Hướng dẫn giải:\(\dfrac{3x-1}{x^2+6x+9}=3.\dfrac{x-\frac{1}{3}}{\left(x+3\right)^2}=3.\left[\dfrac{x+3-\frac{10}{3}}{\left(x+3\right)^2}\right]\)
\(=\dfrac{3}{x+3}-\dfrac{10}{\left(x+3\right)^2}\)
Vậy \(\int\limits^1_0\dfrac{\left(3x-1\right)\text{d}x}{x^2+6x+9}=\int\limits^1_0\dfrac{3}{x+3}\text{d}x-\int\limits^1_0\dfrac{10}{\left(x+3\right)^2}\text{d}x\)
\(=3.\ln\left(x+3\right)|^1_0-10.\dfrac{1}{-2+1}.\left(x+3\right)^{-2+1}|^1_0\)
\(=\left[3\ln\left(x+3\right)+\dfrac{10}{\left(x+3\right)}\right]|^1_0\)
\(=3\left(\ln4-\ln3\right)+10\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\)
\(=3\ln\dfrac{4}{3}-\dfrac{5}{6}\).
Suy ra \(a=4;b=3\) . Từ đó \(ab=12\).