Question

$\int\limits^2_1 {\frac{3x-1}{x+1} } \, dx$

Answer 1

$\displaystyle \int_{1}^2\dfrac{3x-1}{x+1}\, dx = \int_{1}^2\dfrac{3(x+1)-4}{x+1}\, dx =\\ \\ = \int_{1}^2 \dfrac{3(x+1)}{x+1}\, dx -\int_{1}^2 \dfrac{4}{x+1}\, dx =\\ \\ = \int_{1}^2 3\, dx - 4\int_{1}^2 \dfrac{1}{x+1}\, dx = \left(3x\right)\big|_{1}^2-\left(4\ln|x+1|\right)\big|_{1}^2=\\ \\ = (3\cdot 2) - 4\ln|2+1| - 3\cdot 1 + 4\ln|1+1| =\\ \\ = 6 - 4\ln 3 - 4\ln 2 = 6 - 4(\ln 3 - \ln 2)=\\ \\ =\boxed{\boxed{3 -4\ln \dfrac{3}{2}}}$