Integral of x*arctg(4x) dx

1. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = \operatorname{atan}{\left(4 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$.

   Then $\operatorname{du}{\left(x \right)} = \frac{4}{16 x^{2} + 1}$.

   To find $v{\left(x \right)}$:

   1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int x\, dx = \frac{x^{2}}{2}$$

   Now evaluate the sub-integral.

2. The integral of a constant times a function is the constant times the integral of the function:

   $$\int \frac{2 x^{2}}{16 x^{2} + 1}\, dx = 2 \int \frac{x^{2}}{16 x^{2} + 1}\, dx$$

   1. Rewrite the integrand:

      $$\frac{x^{2}}{16 x^{2} + 1} = \frac{1}{16} - \frac{1}{16 \cdot \left(16 x^{2} + 1\right)}$$

   2. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int \frac{1}{16}\, dx = \frac{x}{16}$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- \frac{1}{16 \cdot \left(16 x^{2} + 1\right)}\right)\, dx = - \frac{\int \frac{1}{16 x^{2} + 1}\, dx}{16}$$

         1. The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(4 x \right)}}{4}$.

         So, the result is: $- \frac{\operatorname{atan}{\left(4 x \right)}}{64}$

      The result is: $\frac{x}{16} - \frac{\operatorname{atan}{\left(4 x \right)}}{64}$

   So, the result is: $\frac{x}{8} - \frac{\operatorname{atan}{\left(4 x \right)}}{32}$

3. Add the constant of integration:

   $$\frac{x^{2} \operatorname{atan}{\left(4 x \right)}}{2} - \frac{x}{8} + \frac{\operatorname{atan}{\left(4 x \right)}}{32}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \operatorname{atan}{\left(4 x \right)}}{2} - \frac{x}{8} + \frac{\operatorname{atan}{\left(4 x \right)}}{32}+ \mathrm{constant}$$