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