Integral of $$$\operatorname{atan}{\left(3 x \right)}$$$
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Find $$$\int \operatorname{atan}{\left(3 x \right)}\, dx$$$.
Solution
Let $$$u=3 x$$$.
Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.
Thus,
$${\color{red}{\int{\operatorname{atan}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{3} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\operatorname{atan}{\left(u \right)} d u}}{3}\right)}}$$
For the integral $$$\int{\operatorname{atan}{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$.
Let $$$\operatorname{\omega}=\operatorname{atan}{\left(u \right)}$$$ and $$$\operatorname{dv}=du$$$.
Then $$$\operatorname{d\omega}=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du=\frac{du}{u^{2} + 1}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).
Therefore,
$$\frac{{\color{red}{\int{\operatorname{atan}{\left(u \right)} d u}}}}{3}=\frac{{\color{red}{\left(\operatorname{atan}{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u^{2} + 1} d u}\right)}}}{3}=\frac{{\color{red}{\left(u \operatorname{atan}{\left(u \right)} - \int{\frac{u}{u^{2} + 1} d u}\right)}}}{3}$$
Let $$$v=u^{2} + 1$$$.
Then $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (steps can be seen »), and we have that $$$u du = \frac{dv}{2}$$$.
So,
$$\frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{{\color{red}{\int{\frac{u}{u^{2} + 1} d u}}}}{3} = \frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{3}$$
Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$\frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{3} = \frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}}{3}$$
The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{6} = \frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{6}$$
Recall that $$$v=u^{2} + 1$$$:
$$\frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{6} = \frac{u \operatorname{atan}{\left(u \right)}}{3} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{6}$$
Recall that $$$u=3 x$$$:
$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{6} + \frac{{\color{red}{u}} \operatorname{atan}{\left({\color{red}{u}} \right)}}{3} = - \frac{\ln{\left(1 + {\color{red}{\left(3 x\right)}}^{2} \right)}}{6} + \frac{{\color{red}{\left(3 x\right)}} \operatorname{atan}{\left({\color{red}{\left(3 x\right)}} \right)}}{3}$$
Therefore,
$$\int{\operatorname{atan}{\left(3 x \right)} d x} = x \operatorname{atan}{\left(3 x \right)} - \frac{\ln{\left(9 x^{2} + 1 \right)}}{6}$$
Add the constant of integration:
$$\int{\operatorname{atan}{\left(3 x \right)} d x} = x \operatorname{atan}{\left(3 x \right)} - \frac{\ln{\left(9 x^{2} + 1 \right)}}{6}+C$$
Answer
$$$\int \operatorname{atan}{\left(3 x \right)}\, dx = \left(x \operatorname{atan}{\left(3 x \right)} - \frac{\ln\left(9 x^{2} + 1\right)}{6}\right) + C$$$A