I had the same problem, except different numbers, got them all right previously but this Q, the ones marked in red are incorrect?? Why?

Let's start with the ranges:

u = 8x - 2

The lower limit was x = 3. Plug that in to find the lower limit for u:

u = 8(3) - 2 = 22

The upper limit was x = 6. Plug that in to find the upper limit for u:

u = 8(6) - 2 = 46

Apparently you had these but the screenshot only shows the first digit of each.

Next calculate du/dx:

d(8x - 2)/dx = 8

If du/dx = 8, then dx = du/8

Replace 8x - 2 with u.

Replace dx with du/8

Original:

9 / (8x - 2)² dx

Substituted:

9 / u² * du/8

= (9/8) 1/u² du

Answer:

integral from 22 to 46 of (9/8) 1/u² du

I'll leave the evaluation to you so you hopefully learn how to do this for yourself next time.

Again, you are not doing your part; you still have unattended questions and to vote BA. See

https://ca.answers.yahoo.com/question/index?qid=20...

We answer your questions. Can you answer ours: 

In the question below, why did you vote BA for the person who got it all wrong?

Are you just being a troll? Because you are sure acting like one...Think about it, be honest & grow up !

https://ca.answers.yahoo.com/question/index?qid=20...

If u = 8x - 2, then du = 8.dx

Hence, dx = du/8

so, ∫ [9/(8x - 2)²] dx => ∫ (9/u²)(du/8)

i.e. (9/8) ∫ (1/u²).du

As this is in terms of u, we need to convert the original limits of x in terms of u

so, with x = 6 we have, u = 8(6) - 2 = 46

with x = 3 we have, u = 8(3) - 2 = 22

:)>

I think the definite integral becomes  ∫ 9/(8u²) du and the new limits should be 22 and 46.