Use integration by parts to find the exact value of: ∫ (upper limit 2, lower limit 1) 4x^3 ln2x ?????????

Let u = ln(2x) and dv = 4x³ dx
Then, du = (1/x) dx and v = x⁴

∫4x³ ln(2x) dx
= ∫ln(2x) * (4x³ dx)
= ∫u dv
= u v - ∫v du
= ln(2x) x⁴ - ∫x⁴ (1/x) dx
= x⁴ ln(2x) - ∫x³ dx
= x⁴ ln(2x) - (x⁴/4) + C

₂
∫4x³ ln(2x) dx
¹
= [2⁴ ln(2*2) - (2⁴/4)] - [1⁴ ln(2*1) - (1⁴/4)]
= [16 ln(4) - 4] - [ln(2) - (1/4)]
= [16 ln(2²) - 4] - [ln(2) - (1/4)]
= 32 ln(2) - 4 - ln(2) + (1/4)
= 31 ln(2) - (15/4)

To make life easy, let z = 2x, then dz = 2dx x^3 = =z^3/8 so

Integral[a,b](4x^3 ln(2x) dx) = integral[a,b](1/4 z^3 ln(z) dz)

Now let u = ln(z) du = dz/z and dv = z^3 dz z = z^4/4 then

Integral[a,b](4x^3 ln(2x) dx) = integral[a,b](1/4 z^3 ln(z) dz) = 1/4*(z^4*ln(z)/4|a,b - integral[a,b](z^3/4 dz))

= 1/4*(b^4*ln(b)/4 - a^4 *ln(a)/4 - b^4/16 + a^4/16)

= b^4*ln(b)/16 - a^4*ln(a)/16 - b^4/64 +a^4/64