<s>[INST] 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 ?
 [/INST] Expanding the square bracket and integrating each term individually gives 
\begin{equation*}
\int (x-\frac{1}{2x})^2dx=\int (x^2+\frac{1}{4x^2}-1)dx=\int x^2dx+\frac{1}{4}\int\frac{1}{x^2}dx-\int dx\\
=\frac{x^3}{3}-\frac{1}{4x}-x+C
\end{equation*}
for a constant $C$. 
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