what is the integral of x^(2)arctan(6x) dx?

I suggest you use integration by parts.

Let x^2 dx = du and v = arctan (6x)
u = x^3/3
dv = 6 dx/[1 + (6x)^2]

The integral is uv - Integral of u dv

To find the integral of x^2 * arctan(6x) dx, we can use the technique of integration by parts. This technique involves expressing the integral as a product of two functions and then applying a specific formula to find the integral.

The formula for integration by parts states that ∫ u * v dx = u * ∫ v dx - ∫ (u' * ∫ v dx) dx, where u and v are functions of x, and u' is the derivative of u with respect to x.

In this case, let's assign u = arctan(6x) and dv = x^2 dx. To find du and v, we need to differentiate u and integrate dv:

du/dx = (1/(1 + (6x)^2)) * 6
=> du = (6/(1 + 36x^2)) dx

∫ dv = ∫ x^2 dx
=> v = (1/3) * x^3

Now, using the formula for integration by parts, we can find the integral:

∫ (x^2 * arctan(6x)) dx
= u * v - ∫ (u' * v) dx
= arctan(6x) * (1/3) * x^3 - ∫ ((6/(1 + 36x^2)) * (1/3) * x^3) dx
= (1/3) * arctan(6x) * x^3 - (2/9) * ∫ (x * (1 + 36x^2)^-1) dx

Now, we need to integrate ∫ (x * (1 + 36x^2)^-1) dx. This can be done using the substitution method.

Let u = 1 + 36x^2, du = 72x dx. Rearranging the equation gives dx = du / (72x).

Substituting this back into the integral, we have:
(2/9) * ∫ (x * (1 + 36x^2)^-1) dx
= (2/9) * ∫ (1/u) * (du / (72x))
= (2/9) * (1/72) * ∫ (1/u) du
= (1/324) * ln|u| + C

Finally, substituting the value of u back into the equation, we have:
(1/324) * ln(1 + 36x^2) + C

So, the integral of x^2 * arctan(6x) dx is (1/3) * arctan(6x) * x^3 - (1/324) * ln(1 + 36x^2) + C, where C is the constant of integration.