## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{x^3}{3}-\dfrac{1}{x}+2x-\dfrac{1}{3}$
Calculate the anti-derivative. Now, $y=\int (x+\dfrac{1}{x})^2 dx=\int (x^2+x^{-2}+2) dx$ We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ and where $C$ is a constant of proportionality. Thus, $y=\dfrac{x^{2+1}}{2+1}+\dfrac{x^{-2+1}}{-2+1}+2+C=\dfrac{x^3}{3}-\dfrac{1}{x}+2x+C$ Here, $y(1)=1 \implies C=-\dfrac{1}{3}$ Hence, $y=\dfrac{x^3}{3}-\dfrac{1}{x}+2x-\dfrac{1}{3}$