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

From U-substitution, I got $u=x-\frac{1}{2x},\quad \dfrac{du}{dx} =1+ \frac{1}{2x^2}$ , and $dx= 1+2x^2 du$

and in the end I come 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 8 down vote accepted

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

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\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}-x-\frac{1}{4x}+C \end{equation*}

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