Problems on Techniques of Integration

Use the Integration by Parts. Set

u &=& \arctan(3 x)\\
dv &=& dx\;.


du &=&\displaystyle \frac{3}{9 x^2 + 1} dx\\
v &=& x\;.


\displaystyle \int_0^{\pi/12} \arctan(3 x)...
... \frac{1}{6}\Big[\ln(9x^2 + 1)\Big]_0^{\pi/12}\cdot


\begin{displaymath}\Big[x \arctan(3 x)\Big]_0^{\pi/12} = \frac{\pi}{12} \arctan\left(\frac{\pi}{4}\right) = \frac{\pi}{12}\end{displaymath}


\begin{displaymath}\Big[\ln(9x^2 + 1)\Big]_0^{\pi/12} = \ln\left(\frac{9 \pi^2}{144} + 1 \right) = \ln\left(\frac{ \pi^2}{16} + 1 \right)\end{displaymath}

we get

\begin{displaymath}\int_0^{\pi/12} \arctan(3 x) dx = \frac{\pi}{12} - \frac{1}{6}\ln\left(\frac{ \pi^2}{16} + 1 \right)\;.\end{displaymath}

Detailed Answer.

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