Answer
$y=(\dfrac{x^3}{3})-(\dfrac{1}{x})+2x-(\dfrac{1}{3})$
Work Step by Step
As we know that $\int x^{a} dx=\dfrac{x^{a+1}}{a+1}+C$
Now, we have $y=\int (x+\dfrac{1}{x})^2 dx=\int x^2+x^{-2}+2 dx$
This implies that $y=\dfrac{x^{(2+1)}}{(2+1)}+\dfrac{x^{(-2+1)}}{(-2+1)}+2+C$
$\implies y=\dfrac{x^3}{3}-(\dfrac{1}{x})+2x+C$
Now, we have $y(1)=1$ and $C=-\dfrac{1}{3}$
Hence, $y=(\dfrac{x^3}{3})-(\dfrac{1}{x})+2x-(\dfrac{1}{3})$
