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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 ?

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$dx$ is not $(1+2x^2)\,du$. – André Nicolas May 18 '13 at 2:44
can you show me the correction please, and thanks William, for pointing that out. – user78324 May 18 '13 at 2:49
$dx=\frac{2x^2}{2x^2+1}\,du$, which doesn't see useful, since when we substitute, everything has to be expressed in terms of $u$. – André Nicolas May 18 '13 at 2:55

2 Answers 2

up vote 13 down vote accepted

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

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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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