## University Calculus: Early Transcendentals (3rd Edition)

$(x+1)^3+c$
Separate the differentials and integrate. $\int 3(x+1)^2dx=\int (\dfrac{y-1}{y})dy$ or, $3\int (x+1)^2dx=\int (1-\dfrac{1}{y})dy$ This implies that $y-\ln y=\dfrac{3(x+1)^3}{3}+c$ Thus, $y-\ln y=(x+1)^3+c$