Q:

Integration $\int \left(x-\frac{1}{2x} \right)^2\,dx $

Evaluate $$\int\!\left(x-\frac{1}{2x} \right)^2\,dx. $$
Using integrating by substitution, I got $u=x-\frac{1}{2x},\quad \dfrac{du}{dx} =1+ \frac{1}{2x^2}$ , and $dx= 1+2x^2 du$. In the end, I came up with the answer to the integral as :
$$\left(\frac{1}{3}+\frac{2x^2}{3}\right)\left(x-\frac{1}{2x}\right)^3.$$
Any mistake ?

A:

Just expand
$$
\left(x-\frac{1}{2x}\right)^2 = x^2 - 1 +\frac{1}{4x^2}
$$
and integrate term by term.

