Chapter 13 - Section 13.4 - The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus - Exercises - Page 999: 39

$2-\ln (3)$

Here, we have: $I= \int_0^{2} \dfrac{x}{x+1} \ dx$ Let us consider that $u=x+1 \implies dx=\ du$ Now, we have $I= \int_0^{2} \dfrac{u-1}{u} \ du$ or, $= \int_0^{2} (1-\dfrac{1}{u}) \ du$ or, $=[u-\ln|u|]_0^2 +C$ or, $=[x+1-\ln |x+1|]_0^2 $ or, $=[x-\ln |x+1|]_0^2 $ or, $=[2-\ln |2+1|]-[0-\ln |0+1|] $ or, $=2-\ln (3)$