Answer
$$\int\frac{(t+1)^2-1}{t^4}dt=-\frac{t+1}{t^2}+C$$
Work Step by Step
$$A=\int\frac{(t+1)^2-1}{t^4}dt$$ $$A=\int\frac{t^2+2t+1-1}{t^4}dt$$ $$A=\int\frac{t^2+2t}{t^4}dt$$ $$A=\int\Big(\frac{1}{t^2}+\frac{2}{t^3}\Big)dt$$ $$A=\int\Big(\frac{1}{t^2}+2t^{-3}\Big)dt$$ $$A=-\frac{1}{t}+\frac{2t^{-2}}{-2}+C$$ $$A=-\frac{1}{t}-t^{-2}+C$$ $$A=-\frac{1}{t}-\frac{1}{t^2}+C$$ $$A=-\frac{t+1}{t^2}+C$$
