Is thatOriginally Posted byTreeMoney

INT.[5x (7x -9)^14]dx ?

If yes, then here is one way of doing it.

Integration by parts:

INT.[u]dv = u*v -INT.[v]du ------***

Let u = 5x

So, du = 5dx

And dv = (7x -9)^14 dx = (7x-9)^14 *(7dx /7) = (1/7)(7x-9)^14 *(7dx)

So, v = (1/7)[(1/15)(7x-9)^15] = (1/105)(7x-9)^15

Hence,

INT.[5x (7x-9)^14]dx

= (5x)(1/105)(7x-9)^15 -INT.[(1/105)(7x-9)^15](5dx)

= (1/21)x(7x-9)^15 -(1/21)INT[(7x-9)^15]dx

= (1/21)x(7x-9)^15 -(1/21){(1/7)INT[(7x-9)^15](7dx)}

= (1/21)x(7x-9)^15 -(1/147){(1/16)(7x-9)^16}

= (1/21)x(7x-9)^15 -(1/2352)(7x-9)^16 -------------answer.

Check,

d/dx {(1/21)x(7x-9)^15 -(1/2352)(7x-9)^16}

= (1/21){[x*15(7x-9)^14 *7] +[(7x-9)^15 *1]} -(1/2352){16(7x-9)^15 *7}

= {5x(7x-9)^14 +(1/21)(7x-9)^15} -(1/21)(7x-9)^15

= 5x(7x-9)^14

Okay.