Integration - u substitution problem (Integration by parts?)

[sg001]

Homework Statement

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

Homework 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.

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 ;))

 

[vela]

Homework Statement

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

Homework 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.

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

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.

 

[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??

Thanks in advance.

 

[sg001]

sorry, error in the answer it should be,

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

 

[vela]

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

 

[sg001]

still don't understand because it looks as if he has integrated once to get u^7 and then again to get u^8? I am lost.

 

[vela]

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?

 

[sg001]

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