Tích phân \(\int\limits^2_{\frac{1}{2}}\dfrac{1-3x}{\left(x+1\right)^2}\text{d}x\) bằng
\(\dfrac{4}{3}-3\ln2\).\(3\ln2\).\(\dfrac{4}{3}\).\(-\dfrac{4}{3}+3\ln2\).Hướng dẫn giải:\(\int\limits^2_{\frac{1}{2}}\dfrac{1-3x}{\left(x+1\right)^2}\text{d}x=\int\limits^2_{\frac{1}{2}}\dfrac{4-3\left(x+1\right)}{\left(x+1\right)^2}\text{d}x\)
\(=4\int\limits^2_{\frac{1}{2}}\dfrac{1}{\left(x+1\right)^2}\text{d}x-3\int\limits^2_{\frac{1}{2}}\dfrac{1}{x+1}\text{d}x\)
\(=4\int\limits^2_{\frac{1}{2}}\left(x+1\right)^{-2}\text{d}\left(x+1\right)-3\int\limits^2_{\frac{1}{2}}\dfrac{1}{x+1}\text{d}\left(x+1\right)\)
\(=\left[4.\dfrac{1}{-2+1}\left(x+1\right)^{-2+1}-3\ln\left(x+1\right)\right]|^2_{\frac{1}{2}}\)
\(=\left[-\dfrac{4}{x+1}-3\ln\left(x+1\right)\right]|^2_{\frac{1}{2}}\)
\(=-\dfrac{4}{3}-3\ln3+\dfrac{8}{3}+3\ln\dfrac{3}{2}\)
\(=\dfrac{4}{3}-3\ln2\).