Integral of 2yarctan(4y)dy?

Question

Integral of 2yarctan(4y)dy?

Answer 1

Integral (  2y arctan(4y) dy )

Try integration by parts.

Let u = arctan(4y).  dv = 2y dy

du = 4/[1 + (4y)^2] dy.  v = y^2

y^2 arctan(4y) - Integral (  4y^2 / (1 + (4y)^2)  dy )

Work from here.

Answer 2

integrate by parts

∫u dv = uv - ∫v du

let u = arctan(4y) ; dv = 2y

du = 4dy/(1+16y²) ; v = y²

substitute:

2y² arctan(4y) - ∫4y²/(1+16y²) dy

long division:

2y² arctan(4y) - ∫1/4 - (1/4)/(1+16y²) dy

2y² arctan(4y) - ∫1/4 - (1/64)/(1/16+y²) dy

recall that ∫du/(a²+u²) = 1/a * arctan(u/a)

2y² arctan(4y) - [(1/4)y - (1/64)/(1/4) arctan(4y)]  + C

2y² arctan(4y) - (1/4)y + (1/8)arctan(4y)  + C

hope it helps!