Answer
$$A=\frac{2u^3}{3}+\frac{9u^2}{2}+4u+C$$
Work Step by Step
$$A=\int(u+4)(2u+1)du$$ $$A=\int(2u^2+9u+4)du$$ From Table 1, $$\int cf(x)dx=c\int f(x)dx$$ $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ Therefore, $$A=2\int (u^2)du+9\int(u)du+\int4du$$
From Table 1, we also get the followings $$\int kdx=kx+C$$ $$\int (x^n)dx=\frac{x^{n+1}}{n+1}$$
Therefore, $$A=\frac{2u^3}{3}+\frac{9u^2}{2}+4u+C$$
