Integral of $$$\operatorname{atan}{\left(4 x \right)}$$$

Find $$$\int \operatorname{atan}{\left(4 x \right)}\, dx$$$.

Let $$$u=4 x$$$.

Then $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{4}$$$.

So,

$${\color{red}{\int{\operatorname{atan}{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{4} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\operatorname{atan}{\left(u \right)} d u}}{4}\right)}}$$

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 »).

The integral can be rewritten as

$$\frac{{\color{red}{\int{\operatorname{atan}{\left(u \right)} d u}}}}{4}=\frac{{\color{red}{\left(\operatorname{atan}{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u^{2} + 1} d u}\right)}}}{4}=\frac{{\color{red}{\left(u \operatorname{atan}{\left(u \right)} - \int{\frac{u}{u^{2} + 1} d u}\right)}}}{4}$$

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 $$$u du = \frac{dv}{2}$$$.

Therefore,

$$\frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{{\color{red}{\int{\frac{u}{u^{2} + 1} d u}}}}{4} = \frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{4}$$

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 $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$\frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{4} = \frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}}{4}$$

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{8} = \frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{8}$$

Recall that $$$v=u^{2} + 1$$$:

$$\frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{8} = \frac{u \operatorname{atan}{\left(u \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{8}$$

Recall that $$$u=4 x$$$:

$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{8} + \frac{{\color{red}{u}} \operatorname{atan}{\left({\color{red}{u}} \right)}}{4} = - \frac{\ln{\left(1 + {\color{red}{\left(4 x\right)}}^{2} \right)}}{8} + \frac{{\color{red}{\left(4 x\right)}} \operatorname{atan}{\left({\color{red}{\left(4 x\right)}} \right)}}{4}$$

Therefore,

$$\int{\operatorname{atan}{\left(4 x \right)} d x} = x \operatorname{atan}{\left(4 x \right)} - \frac{\ln{\left(16 x^{2} + 1 \right)}}{8}$$

Add the constant of integration:

$$\int{\operatorname{atan}{\left(4 x \right)} d x} = x \operatorname{atan}{\left(4 x \right)} - \frac{\ln{\left(16 x^{2} + 1 \right)}}{8}+C$$

$$$\int \operatorname{atan}{\left(4 x \right)}\, dx = \left(x \operatorname{atan}{\left(4 x \right)} - \frac{\ln\left(16 x^{2} + 1\right)}{8}\right) + C$$$A