Problems on Techniques of Integration

The Arctangent function is one of these functions whose derivative is very nice (rational function). So it is a good idea to use the integration by parts technique in which $\arctan(x)$ will be differentiated. We may also use the antiderivative of the function $\arctan(x)$ and evaluate it at the points $0$ and $\pi/12$.

Set

$$\left\{\begin{array}{lll} u &=& \arctan(3 x)\\ dv &=& dx\;. \end{array}\right.$$

Then

$$\left\{\begin{array}{lll} du &=&\displaystyle \frac{3}{9 x^2 + 1} dx\\ v &=& x\;. \end{array}\right.$$

Since

$$\int_a^b u dv = [u v]_a^b - \int_a^b v du\;,$$

we get

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

Since

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

and

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

we get

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

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