Answer
$$A=\ln|x|+2\sqrt x+x+C$$
Work Step by Step
$$A=\int\frac{1+\sqrt x+x}{x}dx$$ $$A=\int\frac{1+x^{1/2}+x}{x}dx$$ $$A=\int(\frac{1}{x}+x^{-1/2}+1)dx$$From Table 1, $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ Therefore, $$A=\int (\frac{1}{x})dx+\int(x^{-1/2})dx+\int1dx$$
From Table 1, we also get that $$\int(\frac{1}{x})dx=\ln|x|+C$$ $$\int (x^n)dx=\frac{x^{n+1}}{n+1}+C$$ $$\int kdx=kx+C$$
Therefore, $$A=\ln|x|+\frac{x^{1/2}}{\frac{1}{2}}+x+C$$ $$A=\ln|x|+2\sqrt x+x+C$$
