Integral of x^3arctg(x)dx dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = \operatorname{atan}{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = x^{3}$.

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

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

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

      $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

   Now evaluate the sub-integral.

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

   $$\int \frac{x^{4}}{4 \left(x^{2} + 1\right)}\, dx = \frac{\int \frac{x^{4}}{x^{2} + 1}\, dx}{4}$$

   1. Rewrite the integrand:

      $$\frac{x^{4}}{x^{2} + 1} = x^{2} - 1 + \frac{1}{x^{2} + 1}$$

   2. Integrate term-by-term:

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

         $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

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

         $$\int \left(-1\right)\, dx = - x$$

      PiecewiseRule(subfunctions=[(ArctanRule(a=1, b=1, c=1, context=1/(x**2 + 1), symbol=x), True), (ArccothRule(a=1, b=1, c=1, context=1/(x**2 + 1), symbol=x), False), (ArctanhRule(a=1, b=1, c=1, context=1/(x**2 + 1), symbol=x), False)], context=1/(x**2 + 1), symbol=x)

      The result is: $\frac{x^{3}}{3} - x + \operatorname{atan}{\left(x \right)}$

   So, the result is: $\frac{x^{3}}{12} - \frac{x}{4} + \frac{\operatorname{atan}{\left(x \right)}}{4}$

3. Add the constant of integration:

   $$\frac{x^{4} \operatorname{atan}{\left(x \right)}}{4} - \frac{x^{3}}{12} + \frac{x}{4} - \frac{\operatorname{atan}{\left(x \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{4} \operatorname{atan}{\left(x \right)}}{4} - \frac{x^{3}}{12} + \frac{x}{4} - \frac{\operatorname{atan}{\left(x \right)}}{4}+ \mathrm{constant}$$