Chapter 13 - Section 13.4 - The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus - Exercises - Page 999: 40

$4-4\ln (3)$

Here, we have: $I= \int_{-1}^{1} \dfrac{2x}{x+2} \ dx$ Let us consider that $u=x+2 \implies dx=\ du$ Now, we have $I= 2 \int_{-1}^{1} \dfrac{u-2}{u} \ du$ or, $=2 \int_{-1}^{1} (1-\dfrac{2}{u}) \ du$ or, $=[2u-4\ln|u|]_{-1}^1 +C$ or, $=[2(x+2)-4\ln|(x+2)|]_{-1}^1 +C $ or, $=[2(1+2)-4\ln|(1+2)|]-[2(-1+2)-4\ln|(-1+2)|]$ or, $=6-4 \ln (3)-2+4 \ln |1| $ or, $=4-4\ln (3)$