Answer
$$A=\frac{x^3}{3}+x+\tan^{-1}x+C$$
Work Step by Step
$$A=\int(x^2+1+\frac{1}{x^2+1})dx$$ From Table 1, $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ Therefore, $$A=\int (x^2)dx+\int1dx+\int(\frac{1}{x^2+1})dx$$
From Table 1, we also get that $$\int (x^n)dx=\frac{x^{n+1}}{n+1}+C$$ $$\int kdx=kx+C$$ $$\int(\frac{1}{x^2+1})dx=\tan^{-1}x+C$$
Therefore, $$A=\frac{x^3}{3}+x+\tan^{-1}x+C$$
