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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