# Homework Help: Cant do this integral sub

1. Sep 28, 2010

### percs

integral of (x+25)(x/4+6)^7 dx

no foiling out the power

i thought mayeb putting it as (x+25)((x/4+6)^2)^3.5 dx
and then sub : u = x^2/16 + 3x +36 and du = x/8 + 3 dx and 8du = x +24 dx

but its x+25 that we got ... so yea

2. Sep 28, 2010

### Dick

What's wrong with u=x/4+6? Can't you express x+25 in terms of u?

3. Sep 28, 2010

### fzero

There's a much more obvious substitution that turns the integrand into a sum of monomials. Make the denominator (almost) as simple as possible.

4. Sep 28, 2010

### percs

well then du = dx ... umm how could that be correct?

5. Sep 28, 2010

### Inferior89

If u = x/4 +6 then du is not equal to dx.

6. Sep 28, 2010

### percs

well if u = x/4 +6 then the closest to x+25 is 4u = x+24

7. Sep 28, 2010

### percs

sorry 4du = dx

8. Sep 28, 2010

### Inferior89

The point is to get rid of all the x. You don't need it to have the form c*u where c is some constant.

9. Sep 28, 2010

### Dick

i) du isn't equal to dx and ii) how do you know it couldn't work if you haven't tried it?

10. Sep 28, 2010

### percs

would i be able to do that with integration by parts?

11. Sep 28, 2010

### percs

i did it gives me u = x/4 + 6 and du = dx/4

dont know where to go form there

12. Sep 28, 2010

### fzero

u = x/4 + 6 is the same as x = 4u - 24. Rewrite the entire integral in terms of u and du.

13. Sep 28, 2010

### percs

ok well i get
4* integral (4u-1)((4u-2)/4)^7 du
1/4^6 * integral (4u-1)(4u-2)^7 du

and now im stuck ... once again -_-

14. Sep 28, 2010

### Inferior89

You have:

x/4 + 6 = u (the substitution we did to make the denominator nice)
x + 25 = 4u + 1
dx = 4 du

Use this.

15. Sep 28, 2010

### fzero

You have an algebra mistake in your substitution. u = x/4 + 6 and x = 4u - 24 mean that

(x+25)(x/4+6)^7 dx = ( 4u -24 + 25) (u)^7 ( 4 du)

16. Sep 28, 2010

### percs

ya so i have 4* integral (4u+1)(u^7) du

17. Sep 28, 2010

### percs

so its equal to :

4[ (4(x/4+6)^9)/9 + ((x/4+6)^8)/8] + c ?