Biết $\int\limits_{0}^{1}{\dfrac{x\text{dx}}{x+1}}=a+b\ln 2$ (với...

$\int\limits_{0}^{1}{\dfrac{xd\text{x}}{x+1}}=\int\limits_{0}^{1}{\left( 1-\dfrac{1}{x+1} \right)d\text{x}}=\left. \left( x-\ln \left| x+1 \right| \right) \right|_{0}^{1}=1-\ln 2\Rightarrow a=1;b=-1\Rightarrow a-2b=3$.