How to solve/integrate ∫ x (5x-1)6 dx by parts

The instruction says integrate the problem using "by parts".

In these types of text editors, use "^" to mean "raised to the power of."
∫ x(5x - 1)^6 dx
u = x
du = dx
dv = (5x - 1)^6 dx
v = (1/35)(5x - 1)^7
∫ x(5x - 1)^6 dx =
x(1/35)(5x - 1)^7 - ∫ (1/35)(5x - 1)^7 dx =
(x/35)(5x - 1)^7 - (1/40)(1/35)(5x - 1)^8 + c =
(x/35)(5x - 1)^7 - (1/1400)(5x - 1)^8 + c

Given:

∫[x(5x - 1)^6]dx

∫[v]du = uv - ∫[u]dv

let v = x, then dv = dx
let du = (5x - 1)^6 then u = ∫[(5x - 1)^6]dx

which I will do by substituting letting z = 5x - 1 then dz = 5dx or dx = (1/5)dz

u = (1/5)∫[(z)^6]dz = (1/35)z^7 = (1/35)(5x - 1)^7

∫[x(5x - 1)^6]dx = (x)(1/35)(5x - 1)^7 - (1/35)∫[(5x - 1)^7]dx

Once again let z = 5x - 1, then we know that dx = (1/5)dz

∫[x(5x - 1)^6]dx = (x)(1/35)(5x - 1)^7 - (1/175)∫[(z)^7]dz

∫[x(5x - 1)^6]dx = (x)(1/35)(5x - 1)^7 - (1/1400)z^8 + C

Reversing the substitution for z:

∫[x(5x - 1)^6]dx = (x/35)(5x - 1)^7 - (1/1400)(5x - 1)^8 + C

Choose: u= 5x-1 with a du=5dx
dv=xdx with a v=.5x^2

Then follow the formula.... uv- ∫ vdu
(5x-1)(.5x^2)- ∫5(.5x^2)dx
=6[(5x-1)(.5x^2) - (5/6)x^3] +c