Calculus - Calculate the indefinite integral problem. PLEASE help!?

Question

Evaluate the indefinite integral:

x^2arctan(6x)dx

can anyone show me how to solve this problem please?

Answer 1

Hi, sure, good question.

Let's use Integration by Parts. Use "LIATE" to choose "u", log > inverse trig > algebraic > trig > exponential

u = arctan(6x) ; du = 6 / (36x² + 1) dx

dv= x²dx ; v = x³/3

∫udv = uv - ∫ vdu

(1/3)x³arctan(6x) - ∫ x³/3 * 6/ (36x² + 1)

Looking to the integral: ∫ -6x³/ (108x² + 3). This is improper, because the tops degree is larger than the bottoms. Let's divide this, to yield:

-x/18 + x/18(36x²+ 1)

So we can integrate these separately:

∫-1/18x  dx = -x²/36

∫ x / (36x²+ 1) dx

let u = 36x²+ 1

du = 72x

1/72 ∫ 72 x / 18(36x²+ 1)

= 1/72 * 1/18 ∫ du/u

= 1/1296 ln |36x²+ 1| + C

So that the final answer is, putting all of them together:

(1/3)x³arctan(6x) - x²/36 + 1/1296 ln |36x²+ 1| + C