# Integration - u substitution problem (Integration by parts?)

1. Feb 18, 2012

### sg001

1. The problem statement, all variables and given/known data

Find the integral of 3x* (2x-5)^6*dx, let u= 2x -5.

2. Relevant equations
Im not sure if i am meant to use integration by parts or not?? I was able to do previous questions of the topic just using u sub to get rid of the first x variable.

3. The attempt at a solution
So i started by
u= 2x-5
du/dx = 2
dx= du/2

integral of 3x * (u)^6* du/2
3x/2 *1/7 (u)^7 +c

= 3x/14 * (2x-5)^7 +c
However i know this is wrong because i took the integral in the form of a product (with two x variables).

Please help i have been teaching this topic to myself via online videos as I do not go to school, and want to catch up before i start uni (majoring in physics ;))

2. Feb 18, 2012

### vela

Staff Emeritus
You're fine up to here, but you need to get the integrand in terms of just u, so solve u=2x-5 for x and substitute the result into the integrand.

3. Feb 18, 2012

### sg001

Thanks for the quick reply, although

doing this I get,
x=(5+u)/2
Hence, 3*(5+u/2)*1/2*1/7*(u)^7
= 15/28 *(2x-5)^7 + c

Butthe apparent answer is not this
it is... 3/32 * (2x-5)^8 + 15/28 (2x-5) +c

So i got the last part,, but where does the first part come into the equation??

4. Feb 18, 2012

### sg001

sorry, error in the answer it should be,

3/32 * (2x-5)^8 + 15/28 (2x-5)^7 +c

5. Feb 18, 2012

### vela

Staff Emeritus
You have to substitute for x first, multiply everything out, and then integrate.

6. Feb 18, 2012

### sg001

still dont understand because it looks as if he has integrated once to get u^7 and then again to get u^8? im lost.

7. Feb 18, 2012

### vela

Staff Emeritus
You need to take it one step at a time. You got to
$$\frac{3}{2} \int x u^6\,du$$ which you can't integrate yet. So you substitute for x and get
$$\frac{3}{4} \int (u+5) u^6\,du$$ Now how do you integrate that?

8. Feb 18, 2012

### sg001

just got that ooooohhh moment.
thanks so much for the help greatly appreciated.
live long and prosper ;)