Aloha :)
$$I_1=\int\frac{(1+x)^2}{x(1+x^2)}\,dx=\int\frac{1+2x+x^2}{x(1+x^2)}\,dx=\int\left(\frac{2x}{x(1+x^2)}+\frac{1+x^2}{x(1+x^2)}\right)dx$$$$\phantom{I_1}=\int\left(\frac{2}{1+x^2}+\frac{1}{x}\right)dx=2\arctan(x)+\ln|x|+C$$
$$I_2=\int\frac{\sqrt x-1}{x}\,dx=\int\left(\frac{\sqrt x}{x}-\frac1x\right)dx=\int x^{-\frac12}\,dx-\int\frac1x\,dx=\frac{x^{\frac12}}{\frac12}-\ln|x|+C$$$$\phantom{I_2}=2\sqrt x-\ln|x|+C$$
$$I_3=\int\frac{x^4}{1+x^2}\,dx=-\int\frac{-x^4}{1+x^2}\,dx=-\int\frac{1-x^4-1}{1+x^2}\,dx=-\int\left(\frac{1-x^4}{1+x^2}-\frac{1}{1+x^2}\right)dx$$$$\phantom{I_3}=-\int\left(\frac{\cancel{(1+x^2)}(1-x^2)}{\cancel{1+x^2}}-\frac{1}{1+x^2}\right)dx=-\int(1-x^2)dx+\int\frac{1}{1+x^2}dx$$$$\phantom{I_3}=\frac{x^3}{3}-x+\arctan(x)+C$$