Find integral of yArctan(2y)dy (having a hard time breaking it down and really understanding the integration.

Question

Answer 1

Use integration by parts:

Let u = Arctan(2y). Then du = [2 / (1 + 4y²)]dy

Let dv = ydy. Then v = (1/2)y².

So, ∫yArctan(2y)dy = (1/2)y²Arctan(2y) - ∫ [y² / (1 + 4y²)]dy

Use long division to get (1/2)y²Arctan(2y) - ∫ [1/4 - (1/4)(1 / (1 + 4y²)]dy

∫yArctan(2y)dy = (1/2)y²Arctan(2y) - (1/4)y + (1/8)Arctan(2y) + C