Integral of xarctan(6x)dx?

Question

Integral of xarctan(6x)dx?

Answer 1

Start off with integration by parts.

Let u = arctan(6x), dv = x dx, v = (1/2)x^2, and du = [6/(1 + 36x^2)]dx.

uv - ∫v du

(1/2)x^2 arctan(6x) - 3∫x^2/(1 + 6x^2) dx

For the second integral, you will have to apply polynomial long division (which gives you (1/6) + [1/(36x^2 + 6)].

(1/2)x^2 arctan(6x) - 3∫[(1/6) - [1/(36x^2 + 6)])] dx

The second term in the integral requires an inverse trig substitution (since it's in the form du/(u^2 + a^2), where a = √6 and u = 6x.  Answers to integrals like these are (1/a)arctan(u/a) + C).  Since u = 6x, du = 6dx, and the solution to that integral is (√6/36)arctan[(6x√6)/6] + C (with the denominators rationalized, though that step is optional).

The final answer to your original integral, fully simplified, is:

(1/2)x^2arctan(6x) - (1/2)x - (√6/12)arctan(x√6) + C

Answer 2

((36x^2+1)tan^-1(6x)-6x)/72

Answer 3

lets try integration by parts

u= arctan(6x)

du = 6 / radical(1+x^2) dx

dv= x dx

v = 1/2 * x^2

uv - integral(vdu)

1/2 * x^2 * arctan(6x) - integral(1/2 * x^2 * 6 / radical(1+x^2))

solve for that integral

integral ( 3x^2 / radical(1+x^2) dx)

and go from there