Question

Cum se rezolva aceasta integrala? Modulul in mod special​

Answer 1

$\Big|\dfrac{x-2}{x+1}\Big| = \left\{\begin{array}{II} \dfrac{2-x}{x+1},\quad x\in[1,2) \\\\ \dfrac{x-2}{x+1},\quad x\in [2,3]\end{array}\right \\ \\ \\ \displaystyle \int_1^3 \Big|\dfrac{x-2}{x+1}\Big| \, dx = \int_{1}^2\dfrac{2-x}{x+1}\, dx + \int_{2}^3\dfrac{x-2}{x+1}\, dx =$

$\displaystyle=-\int_{1}^2\dfrac{x-2}{x+1}\, dx + \int_{2}^3\dfrac{x-2}{x+1}\, dx = \\ \\ = -\int_{1}^2\dfrac{x+1-3}{x+1}\, dx + \int_{2}^3\dfrac{x+1-3}{x+1}\, dx = \\ \\ = -\int_{1}^2 \Big(1-\dfrac{3}{x+1}\Big)\, dx + \int_{2}^3\Big(1-\dfrac{3}{x+1}\Big)\, dx = \\ \\ =-x\Big|_1^2+3\int_1^2 \dfrac{1}{x+1}\, dx +x\Big|_2^3 - 3 \int_2^3 \dfrac{1}{x+1}\, dx = \\ \\ = -2+1+3\ln|x+1|\Big|_1^2+3-2-3\ln|x+1|\Big|_2^3 =$

$=3\ln3-3\ln 2-3\ln 4+3\ln 3 = 3\ln \dfrac{3\cdot 3}{2\cdot 4} = \boxed{3\ln \dfrac{9}{8}}$