Calc1.4

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$\int \arctan(2x)\,dx\,$

Make a substitution $s = 2x, ds=2\,dx\,$ . Then the problem becomes

$\frac{1}{2}\int \arctan(s) ds\,$

Integrate by parts. Let

$u=\arctan(s), \,\,\,\,dv=ds\,$
$du=\frac{1}{s^2+1}ds, \,\,\,\,v=s\,$

$\frac{1}{2}\left[\int u \,dv\right] = \frac{1}{2}\left[uv-\int v\,du\right] = \frac{1}{2}\left[s\cdot\arctan(s) - \int \frac{s}{s^2+1} \,ds\right]\,$

$=\frac{1}{2}\left[ s\cdot\arctan(s) - \frac{1}{2}\ln(s^2+1)\right]\,$

$=\frac{1}{2}\left[ 2x \cdot \arctan(2x) - \frac{1}{2}\ln((2x)^2+1)\right]\,$

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Calc1.4

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