Answer
$$\int\Big(t-\frac{2}{t}\Big)\Big(t+\frac{2}{t}\Big)dt=\frac{t^3}{3}+\frac{4}{t}+C$$
Work Step by Step
$$A=\int\Big(t-\frac{2}{t}\Big)\Big(t+\frac{2}{t}\Big)dt$$
We have $(a-b)(a+b)=a^2-b^2$. Therefore, $$A=\int\Big(t^2-\frac{4}{t^2}\Big)dt=\int(t^2-4t^{-2})dt$$ $$A=\int t^2dt-4\int t^{-2}dt$$ $$A=\frac{t^3}{3}-\frac{4t^{-1}}{-1}+C$$ $$A=\frac{t^3}{3}+\frac{4}{t}+C$$
