`int x(2x + 5)^8 dx` Evaluate the indefinite integral.

Since it would be very hard to raise to the 8th power the binomial 2x + 5, you need to use the following substitution  `2x+5=u` , such that:

`2x+5 = u=>2dx = du => dx= (du)/2`

`x = (u-5)/2`

`int x*(2x+5)^8dx = (1/4) int (u-5)*u^8 du`

`(1/4) int (u-5)*u^8 du = (1/4) int u^9du - (1/4) int 5u^8 du`

`(1/4) int (u-5)*u^8 du = (u^10)/40 - (5u^9)/36 + c`

Replacing back `2x+5` for u yields:

`int...

Since it would be very hard to raise to the 8th power the binomial 2x + 5, you need to use the following substitution  `2x+5=u` , such that:

`2x+5 = u=>2dx = du => dx= (du)/2`

`x = (u-5)/2`

`int x*(2x+5)^8dx = (1/4) int (u-5)*u^8 du`

`(1/4) int (u-5)*u^8 du = (1/4) int u^9du - (1/4) int 5u^8 du`

`(1/4) int (u-5)*u^8 du = (u^10)/40 - (5u^9)/36 + c`

Replacing back `2x+5` for u yields:

`int x*(2x+5)^8dx = ((2x+5)^10)/40 - (5(2x+5)^9)/36 + c`

Hence, evaluating the indefinite integral, yields `int x*(2x+5)^8dx = ((2x+5)^10)/40 - (5(2x+5)^9)/36 + c.`