Question

Integrate `(2x-1)(3x^2-3x+5)^8`?

Answer 1

`1/27(3x^2-3x+5)^9+C`.

Explanation:

Let, `I=int(2x-1)(3x^2-3x+5)^8dx`

Subst. `(3x^2-3x+5)=u :. (6x-3)dx=du, i.e.,`

`(2x-1)dx=1/3du`.

`:. I=intu^8*1/3du`,

`=1/3*u^(8+1)/(8+1)`,

`=1/27*u^9`.

Since, `u=(3x^2-3x+5)`, we have,

`I=1/27(3x^2-3x+5)^9+C`.

Answer 2

`int(2x-1)(3x^2-3x+5)^8dx=1/27(3x^2-3x+5)^9+C`

Explanation:

.

`int(2x-1)(3x^2-3x+5)^8dx`

Let `u=3x^2-3x+5`

`du=(6x-3)dx=3(2x-1)dx`

`dx=(du)/(3(2x-1))`

Let's substitute:

`int(2x-1)(3x^2-3x+5)^8dx=int(2x-1)u^8(du)/(3(2x-1))=intcancelcolor(red)((2x-1))u^8(du)/(3cancelcolor(red)((2x-1)))=int1/3u^8du=1/3intu^8du=1/3*1/9u^9+C`

Now, we can substitute back:

`int(2x-1)(3x^2-3x+5)^8dx=1/27(3x^2-3x+5)^9+C`